Download - Two-Sample Inference Procedures with Means
Two-Sample Inference
Procedures with Means
Of the following situations, decide which should be analyzed using one-sample matched pair procedure and which should be analyzed using two-sample procedures?
A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to a random sample, company researchers take a weight measurement on each person. After a month of using the drug, each person’s weight is measured again.
Matched pair
Of the following situations, decide which should be analyzed using one-sample matched pair procedure and which should be analyzed using two-sample procedures?
A researcher wants to know if a population of brown rats on one city has a greater mean length than a population in another city. She randomly selects rats from each city and measures the lengths of their tails.
Two independent samples
Of the following situations, decide which should be analyzed using one-sample matched pair procedure and which should be analyzed using two-sample procedures?
A researcher wants to know if a new vitamin supplement will make the tails of brown rats grow longer. She takes 50 rats and divides them into 25 pairs matched by gender and age. Within each pair, she randomly selects one rat to receive the new vitamin. After six months, she measures the length of the rat’s tail.
Matched pair
Of the following situations, decide which should be analyzed using one-sample matched pair procedure and which should be analyzed using two-sample procedures?
A college wants to see if there’s a difference in time it took last year’s class to find a job after graduation and the time it took the class from five years ago to find work after graduation. Researchers take a random sample from both classes and measure the number of days between graduation and first day of employment
Two independent samples
Two independent samples
Difference of Means
CONDITIONS:CONDITIONS:1) The two samples are chosen independently of each
other. OR The treatments are randomly assigned to individuals
or objects or (vice versa) 2) The sample distributions for both samples should be
approximately normal- the populations are known to be normal, or
- the sample sizes are large (n 30), or - graph data to show approximately normal3) 10% rule – Both samples should be less than 10% of
their respective populations
Differences of Means (Using Independent Samples)
Confidence intervals:
statistic of SD valuecritical statisticCI
21xx *t
2
2
2
1
2
1
n
s
n
s
Called standard
error
Differences of Means (Using Independent Samples)
Degrees of FreedomOption 1: use the smaller of the
two values n1 – 1 and n2 – 1
This will produce conservative This will produce conservative results – higher p-values & results – higher p-values & lower confidence.lower confidence.
Option 2: approximation used by technology
2
2
2
21
2
1
1
2
2
2
2
1
2
1
11
11
ns
nns
n
ns
ns
df
Calculator does this
automatically!
Hypothesis Statements:
H0: 1 - 2 = hypothesized value
Ha: 1 - 2 < hypothesized value
Ha: 1 - 2 > hypothesized value
Ha: 1 - 2 ≠ hypothesized value
H0: 1 = 2
Ha: 1< 2
Ha: 1> 2
Ha: 1 ≠ 2
Differences of Means (Using Independent Samples)
Hypothesis Test:
statistic of SD
parameter - statisticstatisticTest
t
2
2
2
1
2
1
2121
ns
nsxx
State the degrees of freedom
Differences of Means (Using Independent Samples)
Matched Pairs(Special type of one-
sample means)
CONDITIONS:CONDITIONS:1) The samples are paired. The sample differences can be
viewed as a random sample from a population of differences.
2) The sample distribution of differences is approximately normal
- the populations of differences is known to be normal, or
- the number of sample difference is large (n 30), or
- graph data to show approximately normal
3) 10% rule – The sample of differences is not more than 10% of the population of differences.
Differences of Paired Means (Matched Pairs)
dx *t
1df n
statistic of SD valuecritical statisticCI
ds
n
Differences of Paired Means (Matched Pairs)
Hypothesis Statements:
H0: d = hypothesized value
Ha: d < hypothesized value
Ha: d> hypothesized value
Ha: d ≠ hypothesized value
Differences of Paired Means (Matched Pairs)
Parameter:d = true mean difference in …
Hypothesis Test:
statistic of SD
parameter - statisticstatisticTest
t
Differences of Paired Means (Matched Pairs)
d d
d
xs
n
1df n
Two competing headache remedies claim to give fast-acting relief. An experiment was performed to compare the mean lengths of time required for bodily absorption of brand A and brand B. Assume the absorption time is normally distributed. Twelve people were randomly selected and given an oral dosage of brand A. Another 12 were randomly selected and given an equal dosage of brand B. The length of time in minutes for the drugs to reach a specified level in the blood was recorded. The results follow: mean SD n Brand A
20.1 8.7 12 Brand B18.9 7.5 12
Example 1
Is there sufficient evidence that these drugs differ in the speed at which they enter the blood stream?
Parameters and Hypotheses
μA = the true mean absorption time in minutes for brand A
Assumptions (Conditions)
Since the conditions are met, a t-test for the two-sample means is appropriate.
1) The samples must be random and independent which is stated in the problem.
H0: μA - μB = 0Ha: μA - μB 0
3) The samples should be less than 10% of their populations. The population should be at least 240 people, which I will assume.
2) The sample distributions should be approximately normal. Since it is stated in the problem that the population is normal then the sample distributions are aprroximately normal.
4) A and B are both unknown
μB = the true mean absorption time in minutes for brand B
μA - μB = the true difference in means in absorption times in minutes for brands
A and B
2 ( .3619) .721p value P t
Calculations
.721 .05
20.1Ax
12An
= 0.05
8.7As
0A B
.361921df
18.9Bx
7.5Bs
12Bn
2 2
A B A B
A B
A B
x xt
s s
n n
2 2
20.1 18.9 0
8.7 7.5
12 12
Conclusion:
Decision: Since p-value > , I fail to reject the null hypothesis at the .05 level.
There is not sufficient evidence to suggest that there is a difference in the true mean absorption time in minutes for Brand A and Brand B.
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μA = the true mean time it takes to commute taking Route A
State the parameters
Justify the confidence interval needed (state assumptions)
3) The samples should be less than 10% of the population. The population should be at least 400 days, which I will assume.
2) The sample distributions should be approximately normal. It is stated in the problem that graphs of the travel times are roughly symmetric and show no outliers, so we will assume the distributions are approximately normal.
Since the conditions are satisfied a t – interval for the difference of means is appropriate.
1) The samples must be random and independent which is state in the problem.
4) A and B are both unknown
μB = the true mean time it takes to commute taking Route B
μB - μA = the true difference in means in time it takes to commute taking Route B from Route A
We are 95% confident that the true mean difference between the commute times is between 1.3599 and 4.6401 minutes.
Calculate the confidence interval.
95% CI
Explain the interval in the context of the problem.
40Ax
3As
1.3599,4.6401 33df
43Bx
2Bs
20Bn 20An
2 2B A
B AB A
s sx x t
n n
2 22 343 40
20 20 2.043
Parameters and Hypotheses
μd = the true mean difference in scores between June and August for students who repeated the course in summer school
Ho: μd = 0Ha: μd > 0
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Assumptions (Conditions)
Since the conditions are met, a t-test for the matched pairs is appropriate.
1) The samples must be paired and random. The samples are from the same student so they are paired and we will assume the sample differences are a random sample of the population of differences..
3) The sample should be less than 10% of the population. The population should be at least 60 students, which we will assume.
2) The sample distribution should be approximately normal.
4) is unknown
The normal probability plot is fairly linear and the boxplot shows no outliers, so we will assume that the sample distribution of differences is approximately normal.
( 1.7541) .06988p value P t
Calculations
.06988 .05
dx
6n
= 0.05ds
0x d d x
d
xt
s
n
(0)
6
1.7541
5df
5.333
7.4475
5.333
7.4475
Conclusion:
Decision: Since p-value > , I fail to reject the null hypothesis at the .05 level.
There is not sufficient evidence to suggest that the true mean difference in scores is different from June to August.
This suggests that program may not be worthwhile.