28. A health researcher conducted an experiment in which participants watched a

Question

28. A health researcher conducted an experiment in which participants watched a film that either did or did not include a person being injured because of not wearing a seat belt. A week later, as part of a seemingly different study, these same participants reported how important they thought it was to wear seat belts. The 16 participants who had seen the injury film gave a mean rating of 8.9 with an estimated population standard deviation of 2.1. The 36 participants in the control condition had a mean of 7.0 with an estimated population standard deviation of 2.4. Do these results suggest that seeing a movie with a person being injured due to not wearing a seat belt makes attitudes more positive (higher ratings) toward seat belt usage? (Use the.01 level) A) What type of statistical test should be conducted on this data and why? B) Conduct the five steps of hypothesis testing. C) Figure the confidence limits for the 99% confidence interval. D) Figure the effect size. Is this a small, medium, or large effect?

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu1 = mu2 Vs H1 : mu1 > mu2

where mu1 and mu2 are two population means.

Assume alpha = level of significance = 0.01

Given that,

Here we have given population standard deviations and sample sizes are large so we will use two sample z-test.

We can do two sample z-test in ti-83 calculator.

steps :

STAT --> TESTS --> 2-SampZTest --> ENTER --> Highlight on STATS --> ENTER --> Input all the values --> Select alternative : >mu2 --> ENTER --> Calculate --> ENTER

Test statistic = 2.88

P-value = 0.0019966

P-value < alpha

Reject H0 at 1% level of significance.

Conclusion : There is sufficient evidence to say that seeing a movie with person being injured due to not wearing a seat belt makes attitudes more positive toward seat belt usage.

Now we have to find 99% confidence interval for population mean difference.

Ti-83 steps :

STAT --> TESTS --> 9:2-SampZInt --> ENTER --> Highlight on STATS --> ENTER --> Input all the values --> C-level : 0.99 --> Calculate --> ENTER

99% confidence interval for mu1 - mu2 is (0.1999, 3.6001)

We are 99% confident that population mean difference is lies between 0.1999 and 3.6001.

Now we have to find effect size.

n1 16 Xbar1 8.9 s1 2.1

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