Question

assume that you would like to compare the mean number of hospital visits per person per year between Texas and California.  Specifically you would like to test if Texans have a different mean number of visitations per year per person than Californians.  In order to do this, you take a random sample of 29 Texans and 24 Californians and ask each of them how many times they visited the Emergency room last year.  The average number of visits for the Texans was 1.158 with a sample standard deviation of .022 while the average number of visits for Californians was 1.378 with a sample standard deviation of .035.  You may assume the distribution of visits in both Texas and California are normally distributed with the same standard deviation.  Assume alpha is 0.01.  

1. What is(are) the critical value(s) for this study?
   1. ± 1.674         b. ±2.40         c. 2.40      d. ±1.675   e.  1.675      f. ± 2.672       g.  ± 2.676          h. 2.676    i.   – 1.674

2. Perform the appropriate test and select the proper conclusion.
   1. There is overwhelming evidence (pvalue < .0001) that the mean number of Emergency Room visits in Texas and California are different.  
   2. There is not sufficient evidence to suggest (pvalue = 1.566) that the mean number of Emergency Room visits in Texas and California are different.  
   3. There is overwhelming evidence (pvalue < .0001) that the proportion of Emergency Room visits in Texas and California are different.
   4. There is not sufficient evidence (pvalue = 1.566) that the proportion of Emergency Room visits in Texas and California are different.
   5. There is not sufficient evidence (pvalue < .0001) that the mean number of Emergency Room visits in Texas and California are different.

3. T  /  F  A 90% confidence interval for the true proportion p is wider than a 95% confidence interval when calculated on the same data set.  

4. T  /  F  If we get a P-value of 0.56, this means we have proven the null hypothesis to be true.  

Answer 1