Question

A researcher conducts a study to investigate any effect that watching the news has on one’s level of optimism. One group of participants are randomly assigned to watch a 60-minute recording of a local news episode and another group of participants are randomly assigned to watch a 60-minute biopic covering the life of a famous news anchor. The 11 people in the biopic condition averaged an optimism level of 72 (SD = 11) and the 16 people in the news condition averaged an optimism level of 61 (SD = 18). Use this information to answer the following questions.

1. Create a 99% confidence interval to estimate the difference in optimism when comparing those who participate in the local news condition versus those who participate in the biopic condition. Be sure to properly report your confidence interval in brackets. Also, be sure to show calculations properly per the instructions for credit.

2. Use the correct statistical analysis to compute the calculated test value. Be sure to show calculations properly per the instructions for credit.

3. Regardless of your answer to the previous question, compute an effect size for this study. Be sure to show calculations properly per the instructions for credit.

Answer 1

(1) Confidence Interval for the difference in Means

Given:

News: = 61, s₁ = 18, n₁ = 16,

Biopic: = 72, s₂ = 11, n₂ = 11,

Since s₁/s₂ = 18/11 = 1.6 (it lies between 0.5 and 2) we used the pooled standard deviation

The Pooled Variance is given by:

df = n₁ + n₂ – 2 = 11+ 16 – 2 = 25

The t critical (2 tail) for = 0.01, df = 25 is 2.797

The Confidence Interval is given by () ME, where

( ) = 61 – 72 = -11

The Lower Limit = -11 - 17.18 = -28.18

The Upper Limit = -11 + 17.18 = 6.18

The 99% Confidence Interval is (-28.18 , 6.18)

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(2) Hypothesis for a 2 tailed test

The Hypothesis:

The Test Statistic:

The p Value:   The p value (2 Tail) for t = -1.81, df = 25, is; p value = 0.0836

The Critical Value:   The critical value (2 tail) at = 0.01, df = 25, t critical = -2.787 and +2.787

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 lies in between -2.787 and +2.787, We Fail To Reject H0

Also since P value (0.0836) is > (0.01), We Fail to Reject H0.

The Conclusion: There isn’t sufficient evidence at the 99% significance level to conclude that there is any difference in the level of optimism while watching the news or a biopic.

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(3) The Effect Size given by Cohens d = Absolute difference of means / Pooled SD

d = ABS(61 - 72) / Sqrt(239.42) = 0.71

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