Question

In: Statistics and Probability

A researcher wanted to see whether there was a difference between NYC citizens and Boston citizens...

A researcher wanted to see whether there was a difference between NYC citizens and Boston citizens in the amount of time (in hours) they spent outside during April 2020. He collected two samples – one of NYC citizens and the other of Boston citizens, and measured the amount of time each subject spent outside during April 2020. Sample #1 (of NYC citizens) contained 20 subjects (n1=24), the mean number of hours spent outside was M1=40, and the sample’s variance was S12=64. Sample #2 (of Boston citizens) contained 24 subjects (n2=24), the mean number of hours spent outside was M2=28, and the sample’s variance was S22=25.

a. Which test should be used to analyze your data? Explain your answer. (3 points)

b. Conduct the appropriate nondirectional hypothesis test on these data. Use α=0.05. You must submit your calculations (if you do not submit your calculations, you will lose points). (8 points)

c. Calculate the appropriate effect size measure. You must submit your calculations (if you do not submit your calculations, you will lose points). (4 points)

d. Compute a 95% confidence interval for the mean difference. You must submit your calculations (if you do not submit your calculations, you will lose points). (5 points)

Solutions

Expert Solution

(a) Since we have 2 different populations and we want to know if there is a difference in their means, we use a 2 sample independent t test (with unknown standard deviations)

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(b) Since s1/s2 = 8/ 5 = 1.6 (it lies between 0.5 and 2) we used the pooled variance.

The degrees of freedom used is n1 + n2 - 2 = 20 + 24 - 2 = 42 (since pooled variance is used)

The Hypothesis:

H0:

Ha:

This is a Two tailed test.

The Test Statistic:We use the students t test as population standard deviations are unknown.

The p Value:   The p value (2 Tail) for t = 6.07, df = 42, is; p value = 0.000

The Critical Value:   The critical value (2 tail) at = 0.05, df = 42, t critical = -2.018 and +2.018

The Decision Rule: If t observed is > t critical or If   t observed is < -t critical, Then Reject H0.

Also If the P value is < , Then Reject H0

The Decision: Since t observed (6.07) is > t critical (2.018), We Reject H0.

Also since P value (0.000) is < (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that there is a difference in the mean number of hours spent outside by the citizens of NYC and Boston.

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(c) effect Size = Difference of means / Pooled SD = (40 - 28) / Sqrt(42.64) = 1.83

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(d) The Confidence Interval is given by ( ) ME, where

( ) = 40 – 28 = 12

The Lower Limit = 12 - 3.99 = 8.01

The Upper Limit = 12 + 3.99 = 15.99

The 95% Confidence Interval is (8.01 , 15.99)

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