ARE MY RESPONSES CORRECT??? YES OR NO!!

3.) a.) S = V ( 93.504 ) ? 10 - 1 = 3. 2232 a = 1- 0.99 = 0.01 x/ 2 = 0.005 n-1 = 10 - 1 =9 t = 3. 250 CI = 12.56+ 3.250 (3.2232 = 15. 8726 CI = 12.56- 3.250( 3.2232 = 9.2474 Confidence Interval (9. 2474 , 15.8726 ) b.) Ho: M= 15 ( The mean number of commercial minutes played per hour is equal to 15. H1 : M/ 15 ( The mean number of commercial minutes played per hour is less than 15. C. ) use t- test ( left tailed ) The conditions are : the population should be normal , independent , sample is random, and the population standard deviation is unknown. All of these conditions are met . (d.) t = 12.56-15 = -2.3939 degrees of freedom = 1-1 = 10-1= 9 3.2232 VTO = T. DIST ( - 2. 3939, 9, TRUE ) = 0.0201 test statistics = will P-Value = 0.0201 - 2.3939 E. ) a =.05 since the test statistic is less than the critical Significance level = 0.05 value and falls in the rejection region, we reject Critical = value = - 18331 Ho. Moreover, p-Value _ significance level 0.05, reject Ho. Therefore there is sufficient evidence to conclude that the show plays less than 15 minutes of commercials per hour. The same radio station wants to track the average number of commercial minutes it plays in an hour. A random sample over the past month yielded the following data In minutes: 15.5, 10.2, 8.2, 16, 14.4, 9.5, 8.9, 11.4, 16.3, and 15.2. Compute a 99% confidence interval for the mean number of comm minutes played in an hour. (2 points) ercial The station manager is concerned that listeners will change the station if too The program manager says the show many commercials are played per hour. plays less than 15 minutes of commercials per hour. Set up the hypotheses to test the p ram manager's statement. (1 points)