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Sleep deprivation Does sleep deprivation linger for more than a day? Researchers designed a study using 71volunteer subjects between the ages of 18and 25. All 21participants took a computer-based visual discrimination test at the start of the study. Then the subjects were randomly assigned into two groups. The11subjects in one group, D, were deprived of sleep for an entire night in a laboratory setting. The10subjects in the other group, A, were allowed unrestricted sleep for the night. Both groups were allowed as much sleep as they wanted for the next two nights. On Day 4, all the subjects took the same visual discrimination test on the computer. Researchers recorded the improvement in time (measured in milliseconds) from Day 1to Day 4on the test for each subject. We used Fathom software to randomly reassign the 21subjects to the two groups1000times, assuming the treatment received doesn’t affect each individual’s time improvement on the test. The dot plot shows the approximate randomization distribution of xA-xD

(a) Explain why the researchers didn’t let the subjects choose whether to be in the sleep deprivation group or the unrestricted sleep group.

(b) In the actual experiment, xA-xD=15.92. This value is marked with a blue line in the figure. What conclusion would you draw? Justify your answer with appropriate evidence.

(c) Based on your conclusion in part (b), could you have made a Type I error or a Type II error? Justify your answer

Short Answer

Expert verified

(a) To ensure that the two groups are as similar as feasible and to exclude the effect of any variables that were not examined, the individuals are randomly assigned to the two groups.

(b) The volunteers who were allowed to sleep had a higher mean than the participants who were not allowed to sleep.

(c) Based on our conclusion in part (b), WE have made a Type I error.

Step by step solution

01

Part (a) Step 1: Given information

Subjects in group D=11

Subjects in group A =10

02

Part (a) step 2: Explanation

To ensure that the two groups are as similar as feasible and to exclude the effect of any variables that were not examined, the individuals are randomly assigned to the two groups.

03

Part (b) Step 1: Given information

Subjects in group D=11

Subjects in group A=10

04

Part (b) Step 2: Explanation

We can see that there are nearly no dots above 15.92and to its right on the dot plot.

This suggests that a sample mean difference of 15.92is extremely implausible, and we conclude that the mean appears to be much greater for those who were allowed to sleep than for those who were not.

The volunteers who were allowed to sleep had a higher mean than the participants who were not allowed to sleep.

05

Part (c) Step 1: Given information

Number of subjects in group D=11

Number of subjects in group A=10

06

Part (c) Step 2: Explanation

Find the hypothesis

H0:μ1=μ2

Ha:μ1>μ2

In part(b), we concluded that the means were significantly different and thus the null hypothesis H0was rejected.

Type 1 error: RejectH0, whenH0is true

Type 2 error: Fail to reject H0, when H0is false

Since we rejected the null hypothesis H0, we could have only made a Type I error.

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Explain why the conditions for using two-sample z procedures to perform inference about p1-p2are not met in the settings

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A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of 100students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30of Instructor A’s students and 22of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?

Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one.

State: I want to perform a test of

H0:p1-p2=0

Ha:p1-p2>0

where p1=the proportion of Instructor A's students that passed the state exam and p2=the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use σ=0.05

Plan: If conditions are met, I’ll do a two-sample ztest for comparing two proportions.

Random The data came from two random samples of 50students.

- Normal The counts of successes and failures in the two groups -30,20,22, and 28-are all at least 10.

- Independent There are at least 1000 students who take this driving school's class.

Do: From the data, p^1=2050=0.40and p^2=3050=0.60. So the pooled proportion of successes is

p^C=22+3050+50=0.52

- Test statistic

localid="1650450621864" z=(0.40-0.60)-00.52(0.48)100+0.52(0.48)100=-2.83

- p-value From Table A, localid="1650450641188" P(z≥-2.83)=1-0.0023=0.9977.

Conclude: The p-value, 0.9977, is greater than α=0.05, so we fail to reject the null hypothesis. There is no convincing evidence that Instructor A's pass rate is higher than Instructor B's.

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