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Problem Set "AP REVIEW ASSIGNMENT #6" id:[PSABTPX]

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1) Problem #PRAJWY9 "PRAJWY9 - 257268 - This data is from..."

A) This data is from the respiratory dataset. The data collected reflects the respiratory status of patients

recruited for a randomised clinical multicenter trial. In each of two centres, eligible patients were randomly

assigned to treatment or placebo. During the treatment, the respiratory status (categorised

as poor or good) was determined at each of four, monthly visits. The trial recruited 111 participants (54 in

the treatment group, 57 in the placebo group).

The data below is a two table that represents the effect of treatment on respiratory status on the fourth and

last monthly visit. Does the data support the claim that the treatment type results in a difference in

respiratory status after 4 months?

Which of the following tests would be most appropriate?

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Multiple choice:

Chi-Squared - Goodness of Fit

No, Goodness of Fit compares a hypothesized distribution and a distribution drawn from a sample.

Chi-Squared - Test of Independence

No, the data is drawn from two different populations (placebo or treatment).

Difference of Means

Chi-Squared - Test of Homogeneity

B) Define the null and alternative hypothesis.

A)

Ho: There is no association between respiratory status and treatment

Ha: There is an association between respiratory status and treatment

B)

Ho: There is an association between respiratory status and treatment

Ha: There is no association between respiratory status and treatment

C)

p represents the proportion of the population that have a respiratory status rated as good.

Ho: Pplacebo

= ptreatment

Ha: The proportions differ

D)

p represents the proportion of the population that have a respiratory status rated as good.

Ho: Pplacebo

= ptreatment

Ha: Pplacebo

< ptreatment

Multiple choice:

A

B

C

D

Hints:

Know that "association" and "independent" are synonymous.

C) Refer the two way table, what is the expected count when the observed is 32? Round answer to the

nearest hundredth.

Algebra:

26.70

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Hints:

Expected Count = (row total × column total)/total

Expected Count = (52 × 57)/111 = 26.70

D) Refer the two way table, what is the expected count when the observed is 34? Round answer to the

nearest hundredth.

Algebra:

28.70

Hints:

Expected Count = (row total × column total)/total

Expected Count = (59 × 54)/111 = 28.70

E) What condition or conditions need to be checked before proceeding with the hypothesis test? Click all

that apply.

Check all that apply:

All expected counts are greater than 1

No more than 20% of expected counts are less than 5

No more than 20% of expected counts are more than 5

All expected counts are greater than 10

No more than 50% of expected counts are less than 5

No more than 20% of observed counts are less than 5

Hints:

F) Are all conditions met?

Multiple choice:

Yes

No

Unable to determine with provided information

G) What is the degrees of freedom for this problem?

Algebra:

1

3

df = (r - 1)(c - 1)

110

No, for chi-squared, df does not equal n - 1. df = (# of rows - 1)(# of columns - 1)

Hints:

df = (# of rows - 1)(# of columns - 1) = (2 - 1)(2 - 1) = 1

H) What is the chi-squared test statistic? Round answer to the nearest hundredth.

Algebra:

4.06

4.05

4.07

Hints:


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Using the TI-83:

1) Enter observed counts into L1

2) Enter expected counts into L2

3) Highlight L3, then enter formula: (L1 - L2)2/L2

4) Enter 1-VAR STATS L3.

5) χ2 = Σx VALUE

Using R/Rweb:

Use the following code as a guide:

obs=c(1, 2, 3) ### Enter your observed counts

exp=c(4, 5, 6) ### Enter your expected counts

chi.statistic=sum( (obs-exp)^2 / exp ) ### Calculate your chi-squared statistic

chi.statistic

χ2 = 4.06

I)

Using either the TI-83 or R/Rweb, calculate the p-value. Round answer to the nearest hundredth.

Algebra:

0.04

0.044

Hints:

TI-83:

Click 2nd -> VARS -> X2cdf

χ2cdf(4.06, 10000, 1)

Note:

Use following format for calculating the p-value as chi-squared tests are always right tailed.

χ2cdf( χ2 Value, Large Number (10000 or larger) , degrees of freedom )

R/Rweb:

pchisq(4.06, 1, lower.tail = FALSE)

Note:

Use following format for calculating the p-value:

pchisq( χ2 Value, df , lower.tail = FALSE )

p-value = 0.044 or 0.04 (ROUNDED)

J) Assume that the significance level is 0.05, the best conclusion would be:

A) Reject the null hypothesis.

B) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we

reject the null hypothesis. The data supports the claim that the proportion of individuals that have a

respiratory status of "good" differs between the placebo and treatment populations.

C) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we can

reject the null hypothesis.

D) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we fail

to reject the null hypothesis. The data supports the claim that the proportion of individuals that have a

respiratory status of "good" differs between the placebo and treatment populations.

E) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we fail


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to reject the null hypothesis. The data does not support the claim that the proportion of individuals that have

a respiratory status of "good" differs between the placebo and treatment populations.

F) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we fail

to reject the null hypothesis.

Multiple choice:

A

One must state the conclusion in the context of the problem.

B

C

One must state the conclusion in the context of the problem.

D

One should reject the null hypothesis.

E

One should reject the null hypothesis as the data supports the alternative hypothesis.

F

One should reject the null hypothesis and state the conclusion in the context of the problem.

2) Problem #PRAJNC8 "PRAJNC8 - STATS4STEM.ORG"

A) Instructions on how use technology to calculate the p-value for the t-distribution. You will have to use

either the TI-83 or R/Rweb to complete these Assistment questions as you will need to calculate the exact

p-value. Knowing how to use the t-table will not be good enough to finish this Assistment.

EXAMPLE #1

For df=8 and t=-2:

TI-83 Instructions:

To calculate the p-value using a TI-83, you would use the following procedure:

Click 2nd -> DISTR -> tcd(-10, -2, 8) = 0.0403 ### tcd( LOWER #, HIGHER #, df) -> Use 10 and -10

for higher and lower limits

R/Rweb Instructions:

To calculate the p-value using R/Rweb, you would use the following function:

pt(-2, 8) = 0.0403 ### P(X<t) = pt(t, df)

Here is a shaded density curve that represents the above calculation:

EXAMPLE #2

If df=18 and t=2.3:

TI-83 Instructions:


Click 2nd -> DISTR -> tcd(2.3, 10, 18) = 0.0168 ### tcd( LOWER #, HIGHER #, df) -> Use 10 and

-10 for higher and lower limits




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P(X>2.3) = 1 - P(X<2.3) = 1-pt(2.3, 18) = 0.0168 ### P(X<t) = pt(t, df)

Here is a shaded density curve that represents the above calculation:

Calculate the p-value for a df=23, and t = -2.43. Round answer to the nearest hundredth.

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Algebra:

.01

Hints:

TI-83 Instructions:


Click 2nd -> DISTR -> tcd(-10, -2.43, 23) ### tcd( LOWER #, HIGHER #, df) -> Use 10 and -10 for

higher and lower limits



pt(-2.43, 23) ### P(X<t) = pt(t, df)

Answer is 0.01

B) Calculate the p-value for a df=18, and t=1.14. Round answer to the nearest hundredth.

Algebra:

0.13

Hints:

TI-83 Instructions:


Click 2nd -> DISTR -> tcd(1.14, 10, 18) ### tcd( LOWER #, HIGHER #, df) -> Use 10 and -10 for




P(X>1.14) = 1 - P(X<1.14) = 1-pt(1.14, 18) ### P(X<t) = pt(t, df)

Answer is 0.13

C)

Calculate the p-value for a df=27, and t=0.64. Round answer to the nearest hundredth.

Algebra:

0.26

Hints:

TI-83 Instructions:


Click 2nd -> DISTR -> tcd(0.64, 10, 27) ### tcd( LOWER #, HIGHER #, df) -> Use 10 and -10 for





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P(X>0.64) = 1 - P(X<0.64) = 1-pt(0.64, 27) ### P(X<t) = pt(t, df)

Answer is 0.26

D) Calculate the p-value for a df=14, and t = -1.28. Round answer to the nearest hundredth.

Algebra:

0.11

Hints:

TI-83 Instructions:


Click 2nd -> DISTR -> tcd(-10, -1.28, 14) ### tcd( LOWER #, HIGHER #, df) -> Use 10 and -10 for




pt(-1.28, 14) ### P(X<t) = pt(t, df)

Answer is 0.11

3) Problem #PRAJGXW "PRAJGXW - Hyp Test #10"

In a hypothesis test, which of the following will cause a decrease in β, the probability of making a type II

error?

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Increasing α while keeping the sample size n, fixed

Decreasing the sample size n, while keeping α fixed

Decreasing α while keeping the sample size n, fixed

Increasing the sample size n, while keeping α fixed

4) Problem #PRAJU6V "PRAJU6V - 255520 - The parameter of ..."

A) The parameter of a population is 85. Which of the following sampling distributions are biased?


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A

B

C

D

B) Which sampling distribution is the best estimator of the statistic?

Multiple choice:

A

A and C are both unbiased, however, C is better as it has less variance

B

B is biased.

C

D

D is biased.

5) Problem #PRAEVP5 "PRAEVP5 - 136860 - About 70% of Face..."

A) About 70% of Facebook users are outside the United States. If 100 users are randomly surveyed, find

the probability that:

Do not use the normal approximation to the binomial for these problems (or your answers may not match)

and round this and subsequent answers to the nearest hundredth.

a) At least 75 are from outside the United States.

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Algebra:


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.16

16

Hints:

At least 75 = P(X≥75)

P(X≥75) = 1 - P(X≤74)

TI-83 => 1-binomcdf(100, .7, 74)

R/Rweb => 1-pbinom(74,100,.7)

Type .16

B) b) Exactly 71 are from outside of the United States. Round to nearest hundred.

Algebra:

.09

9

Hints:

"Exactly 71" = P(X=71)

TI-83 => binompdf(100, .7, 71)

R/Rweb => dbinom(71, 100, .7)

Answer = .09 (rounded)

C) c) At most 80 live outside of the United States.

Algebra:

.99

99

Hints:

"at most 80" = P(X≤80)

TI-83

P(X≤80) = binomcdf(100, .7, 80)

R/Rweb

P(X≤80) = pbinom(80,100,.7)

P(X≤80) = 0.9911128

Answer = .99 (ROUNDED)

D) d) Less than 64 live outside of the United States.

Algebra:

.08

8

Hints:

"less than 64" = P(X<64)

TI-83 => binomcdf(100, .7, 63)

R/Rweb => pbinom(63,100,.7)

For the binomial => P(X<64) = P(X≤63)

Answer = .08

E) e) Less than 65 or more than 72 live outside of the United States.

Algebra:

.41

41

.42

42

Hints:

"Less than 65 or more than 72" = P(X<65) + P(X>72)

For the Binomial distribution : P(X<65) + P(X>72) = P(X≤64) + P(X≥73)


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P(X≥73) = 1 - P(X≤72)

P(X≤64)

TI-83 => binomcdf(100, .7, 64)

R/Rweb => pbinom(64,100,.70)

-----------------------------------------

1 - P(X≤72)

TI-83 => 1-binomcdf(100, .7, 72)

R/Rweb => 1-pbinom(72,100,.7)

pbinom(64,100,.70) + (1-pbinom(72,100,.7))

[1] 0.4124448

Answer = .41

6) Problem #PRAJ9EW "PRAJ9EW - 267301 - A simulation was ..."

A) A simulation was run that created 1000 linear regression lines drawn using 1000 random samples of size 10 drawn from a population of interest. The results

of the analysis can be seen in the graphics below:

Write a short synopsis of what you believe the three graphs represent. Make sure to note the difference between the 1 black line and the numerous red lines. In

addition, how do you think the 2nd and 3rd graphs are related to the first? Make sure to use the following words in your answer:

sample, population, alpha, beta

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Ungraded open response:

Hints:

The black line represents the regression line for the population, this line takes the form y-hat = alpha +

beta*x. Each of the red lines are linear regression lines drawn by using random samples drawn from the

population. 1000 random samples were drawn, and subsequently, 1000 regression lines were drawn from

each of the 1000 samples. For each red line, a and b were recorded. The second and third graphs

represent the sampling distributions of a and b. As one can see, a and b vary. However, the center both


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sampling distributions are 1 and we know that a and b are unbiased estimators of alpha and beta.

Therefore, alpha and beta must both be 1.

B) From the graphics, one can tell that:

Multiple choice:

alpha is 1, beta is 1

a is 1, b is 1

No, a and b are statistics that change from sample to sample, alpha and beta are parameters that

remain constant.

alpha is 1, b is 1

a is 1, beta is 1

a, b, alpha, and beta are always 1

No, a and b are statistics that change from sample to sample, alpha and beta are parameters that

remain constant.

C) It is clear from the second and third graphs that a and b are ___________________ estimators of alpha

and beta.

Fill in:

unbiased

Unbiased

UNBIASED

Hints:

Answer = unbiased

7) Problem #PRAJBPP "PRAJBPP - 239673 - The test scores f..."

The test scores for an AP Statistics class have the distribution N(85, 5). How high would a student

need to score to be in the top twenty percent?

Link to z-table.

*** Round answer to the nearest whole number. ***

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Algebra:

89

89.2

.892

89.25

Hints:

z = (x-µ) / σ

The top twenty percent corresponds to a score that is greater than the 80th percentile.

80th percentile corresponds to a z-score of 0.84

0.84 = (x-85)/5

(0.84)(5) + 85 = x

x = 89.2

Answer = 89 (ROUNDED)

8) Problem #PRAJGYJ "PRAJGYJ - 244753 - Twenty-one elasti..."

Twenty-one elastic bands were divided into two groups.


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One of the sets was placed in hot water (60-65 degrees C) for four minutes, while the other was left at

ambient temperature. After a wait of about ten minutes, the amounts of stretch, under a 1.35 kg weight, were

recorded.

Details regarding this dataset can be found here.

Which of the following statements are correct? Check all that apply.

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Heating the bands allows the bands to stretch more than bands left at ambient temperature

The ambient data has a greater IQR than the heated data

The median of the heated data is approximately equal to the 3rd quartile of the ambient data

The heated data has one outlier

If one were to exclude the one outlier, the ambient data is approximately symmetric

The median of the ambient data is greater than the first quartile of the heated data

Hints:

Make sure to read through each option very carefully and check only those that are correct.

9) Problem #PRAKCGD "PRAKCGD - 270230 - If log(x) = 1, ca..."

A) If log(x) = 1, calculate x.

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Algebra:

10

Hints:

If log(x) = 1, then x = 101 = 10

B) If log(y) = 2.5, calculate y. Round answer to the nearest hundredth.

Algebra:


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316.23

316.22

Hints:

If log(x) = 2.5, then x = 102.5 = 316.23 (ROUNDED)

C) If log(y) = 3.3, calculate y. Round answer to the nearest hundredth.

Algebra:

1995.26

Hints:

If log(x) = 3.3, then x = 103.3 = 1995.26 (ROUNDED)


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