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Stat 200 Week 6 Homework

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Lane Chapter 11

18. You choose an alpha level of .01 and then analyze your data.

 

a.      What is the probability that you will make a Type I error given that the null hypothesis is true?

b.     What is the probability that you will make a Type I error given that the null hypothesis is false?

 

 

Lane Chapter 12

7. Below are data showing the results of six subjects on a memory test. The three scores per subject are their scores on three trials (a, b, and c) of a memory task. Are the subjects getting better each trial? Test the linear effect of trial for the data.

a

b

c

4

6

7

3

7

8

2

8

5

1

4

7

4

6

9

2

4

2

a. Compute L for each subject using the contrast weights -1, 0, and 1. That is, compute (-1)(a) + (0)(b) + (1)(c) for each subject.

b. Compute a one-sample t-test on this column (with the L values for each subject) you created.

 

 

13. You are conducting a study to see if students do better when they study all atonce or in intervals. One group of 12 participants took a test after studying forone hour continuously. The other group of 12 participants took a test afterstudying for three twenty minute sessions. The first group had a mean score of75 and a variance of 120. The second group had a mean score of 86 and avariance of 100.

 

a. What is the calculated t value? Are the mean test scores of these two groupssignificantly different at the .05 level?

 

b. What would the t value be if there were only 6 participants in each group? Would the scores be significant at the .05 level?

 

 

Lane Chapter 13

4. Rank order the following in terms of power.

 

 

Population 1      n          Population 2                        Standard

              Mean                               Mean                       Deviation

a                29                        20             43                                 12

b               34                       15             40                                  6

c              105                       24             50                                 27

d              170                        2             120                                10

 

 

 

Illowsky Chapter 9

65. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.

The null and alternative hypotheses are:

 

a. Ho: x ¯ = 4.5, Ha: x ¯ > 4.5

b. Ho: μ ≥ 4.5, Ha: μ < 4.5

c. Ho: μ = 4.75, Ha: μ > 4.75

d. Ho: μ = 4.5, Ha: μ > 4.5

 

71.Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization

thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week theyspend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test,the Type I error is:

 

a. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher

 

b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same

 

c. to conclude that the mean hours per week currently is 4.5, when in fact, it is higher

d. to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

 

 

 

77. An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2.

 

Do the data support your claim at the1% level?

 

 

 

Illowsky Chapter 10

80. At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean differencebetween their jumping and relaxed times would be zero.

Test their hypothesis.

 

Relaxed time (seconds)

Jumping time (seconds)

26

21

47

40

30

28

22

21

 

Relaxed time (seconds)

Jumping time (seconds)

23

25

45

43

37

35

29

32

 

 

91. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 poundswith a standard deviation of 12 pounds. The liquid diet group had amean weight loss of 45 pounds with a standard deviation of 14 pounds.

 

 

120. A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She

takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class.

She conducts a hypothesis test. The data are as follows.

 

 

Player 1

Player 2

Player 3

Player 4

Mean score before class

83

78

93

87

Mean score after class

80

80

86

86

 

The correct decision is:

a. Reject H0.

b. Do not reject the H0

 

 

 

 

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[Solved] Stat 200 Week 6 Homework

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Lane Chapter 11 18. You choose an alpha level of .01 and then analyze your data. What is the probability that you will make a Type I error given that the null hypothesis is true? What is the probability that you will make a Type I error given that the null hypothesis is false? Solution: The probability of making a Type I error given that the null hypothesis is true is 0.01. Probability (Type I error, if the null hypothesis is true) = α = 0.01 If the null hypothesis is false, then there is no way of committing a Type I error. Probability (Type I error, if the null hypothesis is false) = 0.00 Lane Chapter 12 7. Below are data showing the results of six subjects on a memory test. The three scores per subject are their scores on three trials (a, b, and c) of a memory task. Are the subjects getting better each trial? Test the linear effect of trial for the data. a b c 4 6 7 3 7 8 2 8 5 1 4 7 4 6 9 2 4 2 a. Compute L for each subject using the contrast weights -1, 0, and 1. That is, compute (-1)(a) + (0)(b) + (1)(c) for each subject. b. Compute a one-sample t-test on this column (with the L values for each subject) you created. Solution: a) The L value is to be calculated using the contrast weights -1, 0, and 1. So, L = -1(a) + 0(b) + 1(c) = c-a. The calculated values are as shown below in the table - a b c L = c-a 4 6 7 3 3 7 8 5 2 8 5 3 1 4 7 6 4 6 9 5 2 4 2 0 b) The mean value of L is - The standard deviation of the L is - Now, to test whether the subjects are getting better each trial, the L value should be positive. It is because if the subjects are getting better each trial, then the difference between the marks in c and marks in a must be greater than zero. So, we need test - Null Hypothesis - H0 : > 0, and Alternate Hypothesis - H1 : 0. Where is the hypothesized value of the mean of the population to which our sample belongs. The test statistic for the required testing is - As there are 6 suubjects, so n =6. so, the degree of freedom = n-1 = 6-1 =5. Remember that this is a one tailed test as we are testing the mean to be on one side only. The p-value for the t statistic value of 1.7 and degree of freedom 5 for a one tailed test is 0.075. That means, the probability of obtaining the values outside the required region is 0.075. Hence it is significant ony below a 0.075 level of confidence. 13. You are conducting a study to see if students do better when they study all atonce or in intervals. One group of 12 participants took a test after studying forone hour continuously. The other group of 12 participants took a test afterstudying for three twenty minute sessions. The first group had a mean score of75 and a variance of 120. The second group had a mean score of 86 and avariance of 100. a. What is the calculated t value? Are the mean test scores of these two groupssignificantly different at the .05 level? b. What would the t value be if there were only 6 participants in each group? Would the scores be significant at the .05 level? Solution: For the given data: Sample Size: n_1 = 12, 〖 n〗_2=12 Sample Mean: (x ) ̅_1=75 ,(x ) ̅_2=86 Variance: 〖σ_1〗^2 = 120,〖〖 σ〗_2〗^2=100 Hypothetical mean: d=0 Standard Error of Estimate: E=√(((n_1-1) 〖σ_1〗^2+(n_2-1) 〖σ_2〗^2)/((n_1+n_2-2) ))×√(((n_1+n_2 ))/(n_1 n_2 )) Or,E=√(((12-1)(120)+(12-1)(100))/((12+12-2) ))×√(((12+12))/(12×12)) Or,E=4...
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