Question

**Answer the following questions and be sure to show your
work:-**

1. Based on a sample of college students, a researcher calculates a 95% confidence interval of the number of drinks per week consumed. The result is 4.5 ± 1 drink. Based on a sample of college students from a different institution, a researcher calculates a 95% confidence interval of the number of drinks per week consumed and finds a result of 8.5 ± 2 drinks. What if anything can you conclude about those two college student population?

2. Three things influence the margin of error in a confidence interval estimate of a population mean: sample size, variability in the population, and confidence level. For each of these quantities separately, explain briefly what happens to the margin of error as that quantity increases??

Answer #1

1)

The 2nd population has a mean number of drinks consumed per week quite higher than the mean number of drinks consumed per week in 1st Population.

Also, the standard error or the standard deviation of the 1st population is lower than the 2nd population.

2)

If n be the sample size, s the standard deviation and z be the z score of required confidence level.

For margin of error E, **E = Z•S/n ^{0.5} =
Z•S•n^{-0.5}**

(I) If n increases, clearly the margin of error decrease.

(ii) If variability increases, then variance and so standard deviation s increases. S and E are directly proportional, so when S increases then E increases.

(iii) If confidence level increases then its corresponding z value also increases, so E increases.

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