assume that you would like to comparethe mean number of hospital visits per person per year betweenTexas and California.  Specifically you would like totest if Texans have a different mean number of visitations per yearper person than Californians.  In order to do this, youtake a random sample of 29 Texans and 24 Californians and ask eachof them how many times they visited the Emergency room lastyear.  The average number of visits for the Texans was1.158 with a sample standard deviation of .022 while the averagenumber of visits for Californians was 1.378 with a sample standarddeviation of .035.  You may assume the distribution ofvisits in both Texas and California are normally distributed withthe same standard deviation.  Assume alpha is0.01.  

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

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

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

1. T  /  F  If we get a P-value of0.56, this means we have proven the null hypothesis to betrue.