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In a random sample of 20 episodes of CSI: Miami, the average number of cheesy lines said was 11.3 with a standard deviation of 3.1. In a separate random sample of 15 episodes of That's So Raven, the average number of cheesy lines said was 17.5 with a standard deviation of 4.8. If the appropriate 2-sided test was performed to compare the average number of cheesy lines in the two TV shows, which of the following is true?

1) The average number of cheesy lines in CSI: Miami is significantly higher than in That's So Raven.
2) The average number of cheesy lines in That's So Raven is significantly higher than in CSI: Miami.
3) There is no significant difference in the average number of cheesy lines between the two TV shows.
4) The sample sizes are too small to draw any conclusions.

1 Answer

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Final answer:

To compare the average number of cheesy lines in CSI: Miami and That's So Raven, a two-sample t-test can be performed. The average number of cheesy lines in That's So Raven is significantly higher than in CSI: Miami.

Step-by-step explanation:

To compare the average number of cheesy lines in the two TV shows, we can perform a Two-Sample t-test. The null hypothesis is that there is no significant difference between the averages, while the alternative hypothesis is that there is a significant difference. We can use a significance level of 0.05.



Step 1: Calculate the test statistic.



t = (mean1 - mean2) / sqrt((var1/n1) + (var2/n2))



Where mean1 and mean2 are the sample means, var1 and var2 are the sample variances, n1 and n2 are the sample sizes.



t = (11.3 - 17.5) / sqrt((3.1^2/20) + (4.8^2/15))



t = -6.2 / sqrt((0.482 + 0.818))



t = -6.2 / sqrt(1.3)



t = -6.2 / 1.14



t ≈ -5.44



Step 2: Determine the critical value.



For a two-sided test at a significance level of 0.05, we need to compare the absolute value of the test statistic to the critical value from the t-distribution with degrees of freedom equal to the smaller of (n1-1) and (n2-1).



Since the degrees of freedom for CSI: Miami is (20-1) = 19 and for That's So Raven is (15-1) = 14, we use the critical value from the t-distribution with 14 degrees of freedom.



The critical value at a 0.05 significance level is approximately 2.145.



Step 3: Make a decision.



Since the absolute value of the test statistic (5.44) is greater than the critical value (2.145), we reject the null hypothesis. This means that there is a significant difference in the average number of cheesy lines between CSI: Miami and That's So Raven.



Therefore, the correct option is 2) The average number of cheesy lines in That's So Raven is significantly higher than in CSI: Miami.

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