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Quiz #12 Tutorial #1

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1. Soft drinks have a lot of carbohydrates in them. A random sample was taken of 20 different brandsof soft drink, and a 95% confidence interval was calculated for the mean carbohydrate content (ingrams) of all brands of soft drink. Some output is shown below.

One-Sample T: carbo

Variable N Mean StDev SE Mean 95% CI

carbo 20 41.6000 5.9947 1.3405 (38.7944, 44.4056)

(a) What value of t∗ would be used for this confidence interval? Obtain the value from Table D.

(b) Verify that Minitab’s confidence interval is correct (that is, calculate it yourself and show thatyou get the same thing).

2. Some mice were trained to run a maze. Some of the mice were white and some were brown. Oneof the aims of the study was to see whether there is a difference in the time taken to run the mazebetween the white and the brown mice. Six mice of each colour were randomly sampled. Someoutput from the analysis is shown below. The times are in seconds.

Two-sample T for white vs brown

N Mean StDev SE Mean

white 6 17.00 4.56 1.9

brown 6 16.83 4.79 2.0

Difference = mu (white) - mu (brown)

Estimate for difference: 0.166667

95% CI for difference: (-6.77,7.11)

(a) Verify that the 95% confidence interval is correct (that is, calculate it yourself and show thatyou get the same thing).

(b) One of the researchers says “these brown mice ran the maze quicker, so I don’t need to calculatea confidence interval: I know the brown mice are quicker”. Explain briefly in simple languagewhy this researcher is not correct.



1. A new variety of alfalfa is grown in six different test plots. The sample mean is 1.6 tons per acre,and the sample SD is 0.434 tons per acre. Some output from an analysis is shown below.

T-Test of the Mean

Test of mu = 2.000 vs mu < 2.000

Variable N Mean StDev SE Mean T P

C1 6 1.600 0.434 0.177 -2.26 0.037

(a) Explain briefly why the t distribution is used to calculate the P-value, rather than the normaldistribution.

(b) Would you reject the null hypothesis at level α = 0.01? Explain briefly.

(c) What do you conclude, in the context of the problem?

2. Some medical schools specialize in research, while others specialize in primary care. Do they havedifferent enrolments on average? Random samples were taken of schools of each type, and theenrolments at each school recorded. A 95% confidence interval for the difference in mean enrolmentsis from −7to247, while a 90% confidence interval is from 13 to 216.

(a) Suppose a two-sided test is done, using the same data, with the null hypothesis that the meanenrolments are the same, and the alternative hypothesis that the mean enrolments are different.What can you say about the P-value of this test, from the confidence intervals above?

(b) Using α = 0.10, can you conclude that there is a difference in mean enrolments? Explain briefly.



1. A researcher takes a sample of 10 sea urchin eggs, and measures the thickness of the cortex of eachegg. The sample mean thickness is 4.26 mm and the sample SD is 1.00 mm. Previous studies havefound a mean thickness of 5 mm; the researcher is searching for evidence that the population meanis not 5 mm. Some output from the analysis is shown below.

T-Test of the Mean

Test of mu = 5.000 vs mu not = 5.000


cortex 10 4.260 0.999 0.316 -2.34 0.044


(b) Would you reject the null hypothesis at level α = 0.01?


2. Some mice were trained to run a maze. Some of the mice were white and some were brown. Oneof the aims of the study was to see whether there is a difference in the time taken to run the mazebetween the white and the brown mice. Six mice of each colour were randomly sampled. The resultsare shown below. The times are in seconds.


white 6 17.00 4.56 1.9

brown 6 16.83 4.79 2.0

(a) Write down suitable null and alternative hypotheses for this problem.

(b) Calculate the two-sample test statistic t.

(c) Obtain a P-value from Table D, as accurately as the table permits.

(d) Using α = 0.05, what conclusion do you draw? Explain briefly.



1. Spinifex pigeons live in the western Australian desert. They mainly eat seeds. In a study, 16 pigeonswere captured, and the dry seeds in the stomach of each bird were weighed. Assume that these birdsare a random sample of all spinifex pigeons. The weight of dry seeds had a sample mean of 1.37grams and a sample SD of 1.03 grams. Calculate a 95% confidence interval for the mean weight ofdry seeds in the stomach of all spinifex pigeons.

2. It has been suggested that male nurses earn more on average than female nurses. To assess this, 16male nurses and 20 female nurses were randomly sampled, with annual salaries as summarized below:

Female Male

Mean 33,750 33,800SD 250 300Sample size 20 16

(a) Calculate a 95% confidence interval for the difference in mean earnings between male and femalenurses.

(b) Does your confidence interval suggest that male nurses do earn more on average than femalenurses? Explain briefly.



1. The mean running time for Broadway shows in New York is 2 hours 12 minutes (132 minutes). Aproducer takes a random sample of 8 shows put on outside New York, and finds a sample mean of2 hours 5 minutes (125 minutes) with a standard deviation of 11 minutes. The producer wants toknow whether the shows put on outside New York have the same or different average running timeas the New York shows. Assume that running times have an approximately normal distribution, anduse α = 0.05.

(a) Write down suitable null and alternative hypotheses.

(b) Calculate the test statistic t.

(c) Obtain a P-value for your test statistic, as accurately as your table permits.

(d) Draw a conclusion from your P-value.

2. A researcher wishes to test the claim that, on average, more juveniles than adults are classified asmissing persons. The researcher obtains data from the last 5 years (which you can think of as arandom sample of “all possible years”). Some output from the analysis of the data is shown.

Two-sample T for juvenile vs adult


juvenile 5 63355 2811 1257

adult 5 35374 2635 1179

Difference = mu (juvenile) - mu (adult)


T-Test of difference = 0 (vs >): T-Value = 16.24 P-Value = 0.000 DF = 7

(a) Verify that Minitab’s calculation of the test statistic is correct (that is, calculate it yourself andshow that you get the same thing).

(b) Using the method learned in class, how many degrees of freedom do we have? (Minitab’s methodis slightly different.) Use Table D to obtain a P-value as accurately as you can.

(c) Is the P-value you obtained from the table consistent with Minitab’s? Explain briefly.

(a) t = (63355− 35374)/√

28112/5 + 26352/5 = 16.24. (b) 5− 1 = 4 (Minitab’s “7” is the “softwaredf”). With 4 df, 16.24 is off the end of the row, so the one-sided P-value is less than 0.0005 (we areon the correct side). (c) “0.000” (or 0 to 3 decimals) is indeed less than 0.0005, so our answer isconsistent with Minitab. (Strictly, we need to check that anything less than 0.0005 rounds to 0 whenyou take it to 3 decimals, which it does, but an observation that 0 is less than 0.0005 is all right.)



1. During a recent year, the average cost of making a movie was $54.8 million. In the same year, arandom sample of 15 action movies was taken, with a sample mean of $62.3 million and a samplestandard deviation of $9.5 million. We wish to assess whether action movies cost more to make thanmovies in general. Use α = 0.01.





2. The assessed value of a home is an important quantity. Cities use the value to determine propertytax, and realtors use the value in setting a selling price. Do real estate appraisers and tax assessorsagree, on average, in their assessments of homes’ values? Random samples of each appraiser typeassessed 10 homes (different homes in each case), with the following results:

Real estate appraisers Tax assessors

Sample mean 283256 288354Sample SD 3256 2341Sample size 10 10







1. A university is trying to find out the mean travel time from home to campus for its students. Asample of 31 students is taken; the sample mean is 19.4 minutes and the sample SD is 9.6 minutes.Assuming that travel times have approximately a normal distribution, calculate a 90% confidenceinterval for the mean travel time for all students.

2. A sample of 10 municipalities was taken. For each municipality, total legal costs were measured ineach of two consecutive years. The 10 differences had sample mean 3.2 (thousand dollars) and samplestandard deviation 19.124 (thousand dollars).

(a) The municipalities differ greatly in size. Explain briefly why looking at the differences is nonethe-less reasonable.

(b) We are testing the null hypothesis that there has been no change in legal costs, against thealternative that there has been some change. Calculate the test statistic for this test.


(d) What do you conclude about legal costs? Explain briefly.



1. The mean salary of mayors of Canadian cities is reported to be $120,000. Mayors’ salaries haveapproximately a normal distribution. To assess whether the mean is correct or incorrect, a journalisttakes a simple random sample of 5 Canadian cities and finds the mayor’s salary for each one. Thesample mean is $132,389 and the sample SD is $12,121. Use α = 0.05.





2. A pollution control expert is concerned that the mean dissolved oxygen content in river water islower below a town than above it. This has environmental consequences because aquatic life will notsurvive if the oxygen content of the water is too low. The inspector gathers evidence for his concernby collecting 5 specimens of river water from above the town and another 5 specimens from belowthe town (labelled above and below in the output below).

Two sample T for above vs below


above 5 5.000 0.158 0.071

below 5 4.860 0.114 0.051

95% CI for mu above - mu below: ( -0.066, 0.346)

T-Test mu above = mu below (vs >): T = 1.61 P = 0.076 DF = 7

(a) Is this a matched-pairs or a two-sample design? Explain briefly.

(b) What do you conclude from the test? (If the test done is inappropriate, briefly explain why.)



1. A random sample was taken of 25 students who own cars. The amount spent on gasoline per weekwas recorded; the sample mean was $21.00 and the sample standard deviation was $5.50.

(a) Find the value of t∗ for a 95% confidence interval (for the population mean) based on this samplesize.

(b) Calculate a 95% confidence interval for the mean amount spent on gasoline per week by studentswho own cars.

2. Do male and female university students spend different amounts of time on volunteer activities?Random samples of 20 female students and 18 male students were collected, and the sample numberof hours of volunteer activities per week were as summarized below:

Male Female

Mean 2.5 3.8SD 1.5 1.8Sample size 18 20







1. A report by the Gallup Poll said that women visit their doctor an average of 5.8 times per year. Aresearcher, having doubts about this report, took a simple random sample of 20 women and recordedthe number of times these women visited their doctor in the last year. Some Minitab output fromthe analysis is below.

One-Sample T: visits

Test of mu = 5.8 vs not = 5.8

Variable N Mean StDev SE Mean 95% CI T P

visits 20 3.85000 2.51888 0.56324 (2.67113, 5.02887) -3.46 0.003

(a) Is this a one-sided or two-sided test? Do you think this is the kind of test this researcher shouldbe doing? Explain briefly.

(b) Calculate the test statistic t, and verify that Minitab’s value is correct.

(c) What P-value do you get from Table D? Verify that Minitab’s P-value is consistent with this.

(d) What conclusion do you draw, at α = 0.01?

2. A professional tax preparer helped (a random sample of) 10 people prepare their tax returns. Themean time spent with each person was 21 minutes, with a sample standard deviation of 5.6 minutes. Avolunteer helped (a random sample of a different) 14 people prepare their tax returns. The volunteerspent a mean time of 27 minutes with each person, with a standard deviation of 4.3 minutes.

(a) Calculate a 90% confidence interval for the difference in the mean times spent by the two people.

(b) Does your confidence interval suggest that there would be a difference in mean time spent if wewere able to look at all people with tax returns?

(c) The volunteer spent a longer time with each person, on average. Do you think this means thatit is better to be helped by a volunteer than a tax professional? Explain briefly.



1. In Toronto’s real estate market, the mean rental for a 2-bedroom condominium unit is $1765 permonth. A company subsidizing new employee housing wants to see whether the mean rental rate fora certain part of the city is the same or different. The company takes a simple random sample of 152-bedroom condominiums, and obtains a sample mean of $1924 and a sample standard deviation of$374. Use α = 0.05.





2. Some medical schools specialize in research, while others specialize in primary care. Do they havedifferent enrolments on average? Random samples were taken of schools of each type, and theenrolments at each school recorded. The results were as follows:

SE

N Mean StDev Mean

research 17 596 163 40

primary care 16 482 179 45







1. The mean family size in Canada is 3.18. A simple random sample of families in a particular schooldistrict was taken. The aim was to see whether mean family size in that school district differed from3.18. Some Minitab output from the analysis is given below. (Note that the school district includesfamilies of 2 people, that is, those with no children.)

One-Sample T: family



family 24 3.91667 1.44212 0.29437 (3.30771, 4.52562) 2.50 0.020

(a) Is this a one-sided or two-sided test? Was the test used appropriate? Explain briefly.

(b) Calculate the test statistic. Did Minitab get the correct value?

(c) Obtain a P-value for your test statistic using Table D. Is it consistent with Minitab’s P-value?

(d) What conclusion do you draw about family sizes in this school district?

2. A company has many factories. There is concern that too many labour hours are being lost becauseof accidents in the factories. For each factory in a sample of 10, a record is kept of the number ofhours lost in the past week. In each factory, an industrial safety program is then run. The numberof labour hours lost in the week following the program is recorded. Some output from an analysis isshown below.

Paired T for h_before - h_after


h_before 10 53.8 32.1 10.1

h_after 10 48.6 31.0 9.8

Difference 10 5.20 4.08 1.29

95% CI for mean difference: (2.28, 8.12)

T-Test of mean difference = 0 (vs > 0): T-Value = 4.03 P-Value = 0.001

(a) Is this a matched-pairs or a two-sample study? Explain briefly.

(b) What do you conclude from the test? (If the test done is inappropriate, explain why.)




T-Test of the Mean

Test of mu = 2.000 vs mu < 2.000


C1 6 1.600 0.434 0.177 -2.26 0.037




2. In the situations below, would you run a matched-pairs experiment and paired t-test, or a two-sampleexperiment and a two-sample t-test? In each case, explain briefly.

(a) Assessing the effectiveness of a drug on reaction times, as measured before and after the exper-iment.

(b) Comparing the effectiveness of two different diets on two separate groups of individuals.

(c) Comparing the effectiveness of two different brands of aspirin (eg. for treating headaches).



1. An average 30-gram cookie is said to contain 150 calories. A food scientist took a simple randomsample of 15 30-gram cookies (of different types) and measured the calorie content of each one. SomeMinitab output from the analysis is below.

One-Sample T: cookies

Test of mu = 150 vs > 150

95%

Lower

Variable N Mean StDev SE Mean Bound T P

cookies 15 178.000 24.187 6.245 167.001 4.48 0.000

(a) Is a one-sided or two-sided test being done here? How can you tell? Explain briefly.

(b) Verify Minitab’s calculation of the test statistic (that is, calculate it yourself and show you getthe same thing).

(c) Minitab says the P-value is very small. What do you get from Table D? Do Minitab and yourtable give consistent results?

(d) What do you conclude, at α = 0.05?

2. The average cost of a movie ticket in New York City is $10.25, while the average cost of a movieticket in London, England is $19.63. Both these averages are sample means, obtained from randomsamples of 25 movie theatres in each city. The sample standard deviations were $2.57 and $3.20,respectively.

(a) Calculate a 99% confidence interval for the difference in mean ticket prices between London andNew York.

(b) What does your confidence interval say about the relative costs of movie tickets in New Yorkand London?

(c) Why do you think movie tickets might be more expensive in London? Explain briefly.



1. In building construction, a particular kind of large metal pin is used. In the past, these pins havehad a mean breaking strength of 42.3. A sample of 5 pins is taken from a batch. The sample meanis 42.164 and the sample SD is 0.260. The aim is to see whether there is any evidence that the meanbreaking strength of pins in this batch differs from the past value. Some output from the analysis isshown below.

T-Test of the Mean



strength 5 42.164 0.260 0.116 -1.17 0.31






(b) Calculate a 90% confidence interval for the mean change in legal costs.

(c) According to your confidence interval, is there any evidence that the mean legal cost haschanged? Explain briefly.

Quiz #12 Tutorial #1 Answers

1. Soft drinks have a lot of carbohydrates in them. A random sample was taken of 20 different brandsof soft drink, and a 95% confidence interval was calculated for the mean carbohydrate content (ingrams) of all brands of soft drink. Some output is shown below.

One-Sample T: carbo

Variable N Mean StDev SE Mean 95% CI

carbo 20 41.6000 5.9947 1.3405 (38.7944, 44.4056)

(a) What value of t∗ would be used for this confidence interval? Obtain the value from Table D.

(b) Verify that Minitab’s confidence interval is correct (that is, calculate it yourself and show thatyou get the same thing).

(a) t∗ = 2.093 with 19 df. (Looking it up in the table is much the fastest way, butfiguring out what t∗ must have been to get the interval above is also OK.) (b) Margin2.093(5.9947)/

√20 = 2.81; interval from 41.6− 2.81 = 38.79 to 41.6 + 2.81 = 44.41, which

agrees with Minitab to the accuracy used. (Using more digits in the calculation is fine; anysmall discrepancy between the calculated values and the ones given by Minitab are OKtoo, probably a result of Table D being less accurate than Minitab’s internal t-table.)

2. Some mice were trained to run a maze. Some of the mice were white and some were brown. Oneof the aims of the study was to see whether there is a difference in the time taken to run the mazebetween the white and the brown mice. Six mice of each colour were randomly sampled. Someoutput from the analysis is shown below. The times are in seconds.

Two-sample T for white vs brown


white 6 17.00 4.56 1.9

brown 6 16.83 4.79 2.0

Difference = mu (white) - mu (brown)


95% CI for difference: (-6.77,7.11)

(a) Verify that the 95% confidence interval is correct (that is, calculate it yourself and show thatyou get the same thing).

(b) One of the researchers says “these brown mice ran the maze quicker, so I don’t need to calculatea confidence interval: I know the brown mice are quicker”. Explain briefly in simple languagewhy this researcher is not correct.

(a) df = 6 − 1 = 5; t∗ = 2.571. Margin 2.571√

4.562/6 + 4.792/6 = 6.94, interval 17 −16.83± 6.94, from −6.77 to 7.11, as calculated by Minitab. (Actually, I faked this, becauseMinitab uses the software df, which is here 9 rather than 5.) (b) There is always samplingvariability, and so with different samples of mice, the results could be different. Or: theconfidence interval includes positive and negative values, so either the white or brown micecould be quicker; it’s impossible to say. Or: the difference in sample means is very small,which could easily be chance.



T-Test of the Mean

Test of mu = 2.000 vs mu < 2.000


C1 6 1.600 0.434 0.177 -2.26 0.037




(a) population sd not known (b) P-value not less than 0.01, so do not reject null hypothesis(c) no evidence that population mean (yield of alfalfa) is less than 2 tons/acre; it could beequal to 2 tons per acre.

2. Some medical schools specialize in research, while others specialize in primary care. Do they havedifferent enrolments on average? Random samples were taken of schools of each type, and theenrolments at each school recorded. A 95% confidence interval for the difference in mean enrolmentsis from −7to247, while a 90% confidence interval is from 13 to 216.

(a) Suppose a two-sided test is done, using the same data, with the null hypothesis that the meanenrolments are the same, and the alternative hypothesis that the mean enrolments are different.What can you say about the P-value of this test, from the confidence intervals above?

(b) Using α = 0.10, can you conclude that there is a difference in mean enrolments? Explain briefly.

(a) The null hypothesis says that the mean difference is 0. 0 is inside the first interval, sothe P-value is greater than 1 − 0.95 = 0.05. But 0 is outside the second interval, so theP-value is less than 1 − 0.90 = 0.10. Therefore the P-value is between 0.05 and 0.10. (b)Since the P-value is less than 0.10, we should reject the null hypothesis and conclude thatthere is a difference in mean enrolments. (If the answer to (a) is wrong, look for consistencyin the answer to (b): if the student concludes that the P-value is bigger than 0.10, the nullhypothesis should not be rejected.) (Look out for students who say the P-value is biggerthan 0.05, therefore we should not reject the null, without looking at α.)


1. A researcher takes a sample of 10 sea urchin eggs, and measures the thickness of the cortex of eachegg. The sample mean thickness is 4.26 mm and the sample SD is 1.00 mm. Previous studies havefound a mean thickness of 5 mm; the researcher is searching for evidence that the population meanis not 5 mm. Some output from the analysis is shown below.

T-Test of the Mean



cortex 10 4.260 0.999 0.316 -2.34 0.044




(a) population sd not known (b) P-value is not smaller than 0.01, so do not reject null (c)no evidence that the population mean cortex thickness differs from 5 (could be equal to 5).

2. Some mice were trained to run a maze. Some of the mice were white and some were brown. Oneof the aims of the study was to see whether there is a difference in the time taken to run the mazebetween the white and the brown mice. Six mice of each colour were randomly sampled. The resultsare shown below. The times are in seconds.


white 6 17.00 4.56 1.9

brown 6 16.83 4.79 2.0





(a) Null: mean times are equal; alternative: mean times different (or in symbols H0 : µ1 =µ2, Ha : µ1 6= µ2, with µ1, µ2 suitably defined). (b) t = (17−16.83)/

√4.562/6 + 4.792/6 =

0.06. (c) Using 6− 1 = 5 df, value off end of line to left, so P-value greater than 2× 0.25 =0.50. (Accurate answer is 0.9447.) (d) P-value not less than 0.05, so do not reject null:there is no evidence that the two colours of mice take different mean times to run the maze.


1. Spinifex pigeons live in the western Australian desert. They mainly eat seeds. In a study, 16 pigeonswere captured, and the dry seeds in the stomach of each bird were weighed. Assume that these birdsare a random sample of all spinifex pigeons. The weight of dry seeds had a sample mean of 1.37grams and a sample SD of 1.03 grams. Calculate a 95% confidence interval for the mean weight ofdry seeds in the stomach of all spinifex pigeons.

15 df, so t∗ = 2.131 and margin is 2.131(1.03)/√

16 = 0.55. Confidence interval from 0.82to 1.92.

2. It has been suggested that male nurses earn more on average than female nurses. To assess this, 16male nurses and 20 female nurses were randomly sampled, with annual salaries as summarized below:

Female Male

Mean 33,750 33,800SD 250 300Sample size 20 16

(a) Calculate a 95% confidence interval for the difference in mean earnings between male and femalenurses.

(b) Does your confidence interval suggest that male nurses do earn more on average than femalenurses? Explain briefly.

(a) df = 16 − 1 = 15; t∗ = 2.131; margin 2.131√

2502/20 + 3002/16 = 199. Differencein means 33750 − 33800 = −50 (or other way around); confidence interval −50 ± 199,from -249 to 149. (b) No: the interval includes both positive and negative values, whichsuggests that the difference could go either way. (Strictly, the interval suggests that thereis no evidence of a two-sided difference.)


1. The mean running time for Broadway shows in New York is 2 hours 12 minutes (132 minutes). Aproducer takes a random sample of 8 shows put on outside New York, and finds a sample mean of2 hours 5 minutes (125 minutes) with a standard deviation of 11 minutes. The producer wants toknow whether the shows put on outside New York have the same or different average running timeas the New York shows. Assume that running times have an approximately normal distribution, anduse α = 0.05.





(a) Null µ = 132, alternative µ 6= 132. (b) t = (125 − 132)/(11/√

8) = −2.46. (c) with14 df and ignoring the minus sign, between 0.01 and 0.02 one-sided, so double that to getbetween 0.02 and 0.04. (d) The P-value is less than 0.05, so there is evidence that showsoutside New York have a different mean running time.

2. A researcher wishes to test the claim that, on average, more juveniles than adults are classified asmissing persons. The researcher obtains data from the last 5 years (which you can think of as arandom sample of “all possible years”). Some output from the analysis of the data is shown.

Two-sample T for juvenile vs adult


juvenile 5 63355 2811 1257

adult 5 35374 2635 1179

Difference = mu (juvenile) - mu (adult)


T-Test of difference = 0 (vs >): T-Value = 16.24 P-Value = 0.000 DF = 7

(a) Verify that Minitab’s calculation of the test statistic is correct (that is, calculate it yourself andshow that you get the same thing).

(b) Using the method learned in class, how many degrees of freedom do we have? (Minitab’s methodis slightly different.) Use Table D to obtain a P-value as accurately as you can.

(c) Is the P-value you obtained from the table consistent with Minitab’s? Explain briefly.

(a) t = (63355− 35374)/√

28112/5 + 26352/5 = 16.24. (b) 5− 1 = 4 (Minitab’s “7” is the “softwaredf”). With 4 df, 16.24 is off the end of the row, so the one-sided P-value is less than 0.0005 (we areon the correct side). (c) “0.000” (or 0 to 3 decimals) is indeed less than 0.0005, so our answer isconsistent with Minitab. (Strictly, we need to check that anything less than 0.0005 rounds to 0 whenyou take it to 3 decimals, which it does, but an observation that 0 is less than 0.0005 is all right.)


1. During a recent year, the average cost of making a movie was $54.8 million. In the same year, arandom sample of 15 action movies was taken, with a sample mean of $62.3 million and a samplestandard deviation of $9.5 million. We wish to assess whether action movies cost more to make thanmovies in general. Use α = 0.01.





(a) Null µ = 54.8, alternative µ > 54.8. I’m suppressing the millions in this question. (b)t = (62.3 − 54.8)/(9.5/

√15) = 3.06. (c) The test statistic is on the correct side, so, with

15− 1 = 14 df, P-value between 0.0025 and 0.005. (d) The P-value is less than 0.01, so wehave evidence that action movies do cost more to make on average.

2. The assessed value of a home is an important quantity. Cities use the value to determine propertytax, and realtors use the value in setting a selling price. Do real estate appraisers and tax assessorsagree, on average, in their assessments of homes’ values? Random samples of each appraiser typeassessed 10 homes (different homes in each case), with the following results:

Real estate appraisers Tax assessors

Sample mean 283256 288354Sample SD 3256 2341Sample size 10 10





(a) Null: means of two assessed values are equal; alternative: not equal (or in sym-bols H0 : µ1 = µ2, Ha : µ1 6= µ2, with µ1, µ2 suitably defined). (b) t = (283256 −288354)/

√32562/10 + 23412/10 = −4.02, or without the negative sign (if the subtraction

was done the other way around). (c) Between 0.001 and 0.0025 one-sided, so between 0.002and 0.005 (accurate value 0.003). (d) P-value less than 0.05, so reject null and concludethat means different: the two kinds of assessor have different mean values.


1. A university is trying to find out the mean travel time from home to campus for its students. Asample of 31 students is taken; the sample mean is 19.4 minutes and the sample SD is 9.6 minutes.Assuming that travel times have approximately a normal distribution, calculate a 90% confidenceinterval for the mean travel time for all students.

t∗ = 1.697 (30 df), so margin is 1.697(9.6)/√

31 = 2.93 and interval is from 16.47 to 22.32.



(b) We are testing the null hypothesis that there has been no change in legal costs, against thealternative that there has been some change. Calculate the test statistic for this test.


(d) What do you conclude about legal costs? Explain briefly.

(a) The differences compare the same municipality in two different years, so even though theactual values for different municipalities might be quite different, the changes (differences)could be quite consistent. (b) t = (3.2− 0)/(19.124/

√10) = 0.53. (c) Using 10− 1 = 9 df,

P-value is greater than 0.25 one-sided, 0.50 two-sided (off the left end of the row). (d) Noevidence of a change in legal costs (or of a difference between the two years).


1. The mean salary of mayors of Canadian cities is reported to be $120,000. Mayors’ salaries haveapproximately a normal distribution. To assess whether the mean is correct or incorrect, a journalisttakes a simple random sample of 5 Canadian cities and finds the mayor’s salary for each one. Thesample mean is $132,389 and the sample SD is $12,121. Use α = 0.05.





(a) Null µ = 120, 000, alternative µ 6= 120, 000 (“assessing whether the mean is incorrect”,so two-sided). (b) t = (132, 389− 120, 000)/(12, 121/

√5) = 2.47. (c) P-value from Table D

with 5−1 = 4 df is between 0.025 and 0.05 one-sided; double this to get a P-value between0.05 and 0.10. (d) The P-value is not less than 0.05, so there is not evidence that thereported mean salary is wrong. (Even though the sample mean is some way from $120,000,the sample size is so small that this is not a statistically significant difference.)

2. A pollution control expert is concerned that the mean dissolved oxygen content in river water islower below a town than above it. This has environmental consequences because aquatic life will notsurvive if the oxygen content of the water is too low. The inspector gathers evidence for his concernby collecting 5 specimens of river water from above the town and another 5 specimens from belowthe town (labelled above and below in the output below).

Two sample T for above vs below


above 5 5.000 0.158 0.071

below 5 4.860 0.114 0.051

95% CI for mu above - mu below: ( -0.066, 0.346)

T-Test mu above = mu below (vs >): T = 1.61 P = 0.076 DF = 7

(a) Is this a matched-pairs or a two-sample design? Explain briefly.

(b) What do you conclude from the test? (If the test done is inappropriate, briefly explain why.)

(a) two-sample (no pairing between above-town and below-town measurements) (b) test isappropriate. P-value of 0.076 is bigger than 0.05, so do not reject null hypothesis of equal(population) means above and below the town: there is no evidence of a difference.


1. A random sample was taken of 25 students who own cars. The amount spent on gasoline per weekwas recorded; the sample mean was $21.00 and the sample standard deviation was $5.50.

(a) Find the value of t∗ for a 95% confidence interval (for the population mean) based on this samplesize.

(b) Calculate a 95% confidence interval for the mean amount spent on gasoline per week by studentswho own cars.

(a) t∗ = 2.064 (24 df). (b) Margin is 2.064(5.5)/√

25 = 2.27, so interval from 21 − 2.27 =$18.73 to 21+2.27 = $23.27. (If the value in (a) is wrong but the answer to (b) is otherwisecorrect, full credit should be given for (b).)

2. Do male and female university students spend different amounts of time on volunteer activities?Random samples of 20 female students and 18 male students were collected, and the sample numberof hours of volunteer activities per week were as summarized below:

Male Female

Mean 2.5 3.8SD 1.5 1.8Sample size 18 20





(a) Null: mean times are equal; alternative: means not equal (2-sided). (or in sym-bols H0 : µ1 = µ2, Ha : µ1 6= µ2, with µ1, µ2 suitably defined). (b) t = (2.5 −3.8)/

√1.52/18 + 1.82/20 = −2.43, or without the negative sign (if the subtraction was

done the other way around). (c) Using 18 − 1 = 17 df, between 0.01 and 0.02 one-sided,therefore between 0.02 and 0.04 two-sided. (Accurate answer is 0.0261.) (d) P-value lessthan 0.05, so reject null: there is evidence that the mean time spent on volunteer activitiesby males and females is different.


1. A report by the Gallup Poll said that women visit their doctor an average of 5.8 times per year. Aresearcher, having doubts about this report, took a simple random sample of 20 women and recordedthe number of times these women visited their doctor in the last year. Some Minitab output fromthe analysis is below.

One-Sample T: visits



visits 20 3.85000 2.51888 0.56324 (2.67113, 5.02887) -3.46 0.003

(a) Is this a one-sided or two-sided test? Do you think this is the kind of test this researcher shouldbe doing? Explain briefly.

(b) Calculate the test statistic t, and verify that Minitab’s value is correct.

(c) What P-value do you get from Table D? Verify that Minitab’s P-value is consistent with this.

(d) What conclusion do you draw, at α = 0.01?

(a) Two-sided (“not =”). The researcher only “has doubts”, so should be looking for anydifference from 5.8, which is what a two-sided test does. (b) t = (3.85−5.8)/(2.519/

√20) =

−3.46. (c) Using 20−1 = 19 df, the P-value should be between 0.001 and 0.0025 one-sided,or between 0.002 and 0.005 two-sided (doubling). Minitab’s 0.003 is indeed between 0.002and 0.005. (d) The P-value is less than 0.01, so there is evidence that the Gallup Poll isincorrect. (A look at the CI suggests that women visit their doctor less than 5.8 timesper year, but the conclusion from this test can only be that the mean number of visits haschanged, because it’s a 2-sided test.)

2. A professional tax preparer helped (a random sample of) 10 people prepare their tax returns. Themean time spent with each person was 21 minutes, with a sample standard deviation of 5.6 minutes. Avolunteer helped (a random sample of a different) 14 people prepare their tax returns. The volunteerspent a mean time of 27 minutes with each person, with a standard deviation of 4.3 minutes.

(a) Calculate a 90% confidence interval for the difference in the mean times spent by the two people.

(b) Does your confidence interval suggest that there would be a difference in mean time spent if wewere able to look at all people with tax returns?

(c) The volunteer spent a longer time with each person, on average. Do you think this means thatit is better to be helped by a volunteer than a tax professional? Explain briefly.

(a) df 10 − 1 = 9, t∗ = 1.833 (for 90% CI). Margin 1.833√

5.62/10 + 4.32/14 = 3.87,difference in means 21−27 = −6 (or 6, if subtraction done other way around). CI −6±3.87,from −9.9 to −2.1 minutes. (Or 2.1 to 9.9 minutes). (b) The confidence interval is all oneside of 0 (does not include 0), so the CI suggests that the difference is real (reproducible),so there would be a difference. (c) Not necessarily. The tax professional would probably beable to see things in a tax return more quickly (errors, places to claim things in a differentway), and so would be able to give a higher quality of help in less time (or, at least,complete the job of helping in a specific way in less time). (There are probably differentways of tackling this; use your judgement.)


1. In Toronto’s real estate market, the mean rental for a 2-bedroom condominium unit is $1765 permonth. A company subsidizing new employee housing wants to see whether the mean rental rate fora certain part of the city is the same or different. The company takes a simple random sample of 152-bedroom condominiums, and obtains a sample mean of $1924 and a sample standard deviation of$374. Use α = 0.05.





(a) Null µ = 1765, alternative µ 6= 1765 (two-sided). (b) t = (1924 − 1765)/(374/√

15) =1.65. (c) Using 15− 1 = 14 df, 1.65 is between 1.345 and 1.761, so the P-value is between0.05 and 0.10 one-sided. Double this to get a P-value between 0.10 and 0.20 for the two-sided test. (d) The P-value is not less than 0.05, so there is no evidence of a differencein mean rental rate in this part of the city. (Even though the sample mean is a long wayfrom 1765, a combination of a small(-ish) sample size and a large sample SD means thatthe observed result is not significant.)

2. Some medical schools specialize in research, while others specialize in primary care. Do they havedifferent enrolments on average? Random samples were taken of schools of each type, and theenrolments at each school recorded. The results were as follows:

SE

N Mean StDev Mean

research 17 596 163 40

primary care 16 482 179 45





(a) Null: mean enrolments are equal; alternative: mean enrolments different (or in sym-bols H0 : µ1 = µ2, Ha : µ1 6= µ2, with µ1, µ2 suitably defined). (b) t = (596 −482)/

√1632/17 + 1792/16 = 1.91. (c) Using 16−1 = 15 df, P-value between 0.025 and 0.05

one-sided, so between 0.05 and 0.10 two-sided. (Accurate answer is 0.0742.) (d) P-valuenot less than 0.05, so do not reject null: there is no evidence that mean enrolments at thetwo types of school are different.


1. The mean family size in Canada is 3.18. A simple random sample of families in a particular schooldistrict was taken. The aim was to see whether mean family size in that school district differed from3.18. Some Minitab output from the analysis is given below. (Note that the school district includesfamilies of 2 people, that is, those with no children.)

One-Sample T: family



family 24 3.91667 1.44212 0.29437 (3.30771, 4.52562) 2.50 0.020

(a) Is this a one-sided or two-sided test? Was the test used appropriate? Explain briefly.

(b) Calculate the test statistic. Did Minitab get the correct value?

(c) Obtain a P-value for your test statistic using Table D. Is it consistent with Minitab’s P-value?

(d) What conclusion do you draw about family sizes in this school district?

(a) Two-sided, because the alternative was “not =”. Since we are trying to find outwhether mean family size “is different” from 3.18, this is the right test to use. (b) t =(3.917 − 3.18)/(1.442/

√24) = 2.50. Minitab was correct. (c) Using 23 df, the P-value

should be very close to 0.01 one-sided, or 0.02 two-sided. 0.02 is what Minitab had. (Acorrect answer based on “between” is also OK: something like “between 2 × 0.005 and2 × 0.02” is a bit strange, but all right.) (d) There is evidence that the mean family sizediffers from 3.18 in this school district.

2. A company has many factories. There is concern that too many labour hours are being lost becauseof accidents in the factories. For each factory in a sample of 10, a record is kept of the number ofhours lost in the past week. In each factory, an industrial safety program is then run. The numberof labour hours lost in the week following the program is recorded. Some output from an analysis isshown below.

Paired T for h_before - h_after


h_before 10 53.8 32.1 10.1

h_after 10 48.6 31.0 9.8

Difference 10 5.20 4.08 1.29

95% CI for mean difference: (2.28, 8.12)

T-Test of mean difference = 0 (vs > 0): T-Value = 4.03 P-Value = 0.001

(a) Is this a matched-pairs or a two-sample study? Explain briefly.

(b) What do you conclude from the test? (If the test done is inappropriate, explain why.)

(a) matched pairs (same 10 factories each time) (b) the test is appropriate (it is correctly amatched-pairs t-test); the P-value is small (smaller than 0.05) so reject the null hypothesisand conclude that the (population) mean number of hours lost has decreased (look atalternative; logically this is the right test to do, since you would expect fewer hours to belost after the program is run).



T-Test of the Mean

Test of mu = 2.000 vs mu < 2.000


C1 6 1.600 0.434 0.177 -2.26 0.037




(a) because the population sd is not known. (b) reject null hypothesis because P-value lessthan 0.05 (c) population mean (yield of alfalfa) is less than 2 tons per acre. (Or “there isevidence that. . . ”.)

2. In the situations below, would you run a matched-pairs experiment and paired t-test, or a two-sampleexperiment and a two-sample t-test? In each case, explain briefly.

(a) Assessing the effectiveness of a drug on reaction times, as measured before and after the exper-iment.

(b) Comparing the effectiveness of two different diets on two separate groups of individuals.

(c) Comparing the effectiveness of two different brands of aspirin (eg. for treating headaches).

(a) Paired: it would be best to measure the same subjects before and after (for a goodcomparison), and there’s no reason why you can’t do this here. (b) Two-sample: from thedescription, there’s no good way of pairing individuals. (c) Probably two-sample, because(for example) you can only give one brand of aspirin to a subject with a headache (youcan’t give both brands for the same subject and the same headache). You might be ableto make a case for paired: for each subject, randomly choose a brand to treat the firstheadache, and then give the other brand for the second headache, if there is one. But youcould end up wasting data this way.


1. An average 30-gram cookie is said to contain 150 calories. A food scientist took a simple randomsample of 15 30-gram cookies (of different types) and measured the calorie content of each one. SomeMinitab output from the analysis is below.

One-Sample T: cookies

Test of mu = 150 vs > 150

95%

Lower

Variable N Mean StDev SE Mean Bound T P

cookies 15 178.000 24.187 6.245 167.001 4.48 0.000

(a) Is a one-sided or two-sided test being done here? How can you tell? Explain briefly.

(b) Verify Minitab’s calculation of the test statistic (that is, calculate it yourself and show you getthe same thing).

(c) Minitab says the P-value is very small. What do you get from Table D? Do Minitab and yourtable give consistent results?

(d) What do you conclude, at α = 0.05?

(a) One-sided: the alternative is µ > 150 (from the “Test of” line). (b) t = (178 −150)/(24.187/

√15) = 4.48. (c) Table D, using 14 df, says the P-value is less than 0.0005 (be-

cause 4.48 is off the end of the line). Minitab’s value is indeed less than 0.0005. (Minitab’svalue is 0 only to 3 decimals, which is what a value less than 0.0005 would round to.) (d)The P-value is less than 0.05, so there is evidence that the mean calorie content of 30-gramcookies is in fact greater than 150.

2. The average cost of a movie ticket in New York City is $10.25, while the average cost of a movieticket in London, England is $19.63. Both these averages are sample means, obtained from randomsamples of 25 movie theatres in each city. The sample standard deviations were $2.57 and $3.20,respectively.

(a) Calculate a 99% confidence interval for the difference in mean ticket prices between London andNew York.

(b) What does your confidence interval say about the relative costs of movie tickets in New Yorkand London?

(c) Why do you think movie tickets might be more expensive in London? Explain briefly.

(a) df = 25 − 1 = 24; t∗ = 2.797. Margin 2.797√

2.572/25 + 3.202/25 = 2.30; interval is10.25 − 19.63 ± 2.30 or −11.68 to −7.08. (b) Average cost of a movie ticket is cheaper inNYC (or more expensive in London) by somewhere between about $7 and $12. (c) Manypossible answers (use your judgement). A simple answer is that the cost of living is higherin England, so that pretty much everything is more expensive in London than in New YorkCity.


1. In building construction, a particular kind of large metal pin is used. In the past, these pins havehad a mean breaking strength of 42.3. A sample of 5 pins is taken from a batch. The sample meanis 42.164 and the sample SD is 0.260. The aim is to see whether there is any evidence that the meanbreaking strength of pins in this batch differs from the past value. Some output from the analysis isshown below.

T-Test of the Mean



strength 5 42.164 0.260 0.116 -1.17 0.31




(a) population sd not known (b) P-value is bigger than 0.05, so do not reject null (c) noevidence that the mean breaking strength of pins in this batch differs from the past value(the population mean could be 42.3).



(b) Calculate a 90% confidence interval for the mean change in legal costs.

(c) According to your confidence interval, is there any evidence that the mean legal cost haschanged? Explain briefly.

(a) The differences compare the same municipality in two different years, so even though theactual values for different municipalities might be quite different, the changes (differences)could be quite consistent. (b) df 10−1 = 9; t∗ = 1.833; margin 1.833(19.124)/

√10 = 11.09;

interval 3.2 ± 11.1, from −7.9 to 14.3. (c) There is no evidence, because the confidenceinterval contains 0 (or contains both positive and negative values). (Even if it didn’t,we’d only be able to say that the P-value was less than 0.10, which is not of itself strongevidence.)

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