HELLO AGAIN... NEED HELP w HW QUESTIONS!

HE PROVIDED EXPLANATIONS (listed below) BUT THEY'RE NOT FOR ME. NEED MORE CLARITY AND DETAILS SO PLS PROVIDE AS MUCH AS POSSIBLE! THANK YOU

14.7 For each of 500 simulated random samples, a statistics class determined a 95% confidence interval for the mean, and it found that only 464 of them contained the mean of the population sampled; the other 36 did not. At the 0.01 level of significance, is there any real evidence to doubt that the method employed yields 95% confidence intervals?14.8 In a random sample of 600 persons interviewed at a baseball game, 157 complained that their seats were uncomfortable. Test the claim that 30% of the persons at the game would feel that way, using the (a) 0.05 level of significance; (b) 0.01 level of significance.14.11 In a random sample of visitors to the Heard Museum in Phoenix, Arizona, 22 of 100 families from New England and 33 of 120 families from California purchased some Indian jewelry in the gift shop. Use the 0.05 level of significance to test thenull hypothesis p1 = p2, that there is no difference between the corresponding population proportions, against the alternative hypothesis pi # pz.14.12 The service department of a Chrysler dealership offers fresh donuts to customers in its waiting room. When it supplied 12 dozen donuts from Bakery A, it found that 96 were eaten completely while the others were partially eaten and discarded. When it supplied 12 dozen donuts from Bakery B it found that 105 were eaten completely while the others were partially eaten and discarded. At the 0.05 level of significance test whether the difference between the corresponding sample proportions is significant.Question 14.? In 50:] random samples,a 95% confidence interval was calculated. I: was found that 4-64 contained Elie population mean, and the other 36 did not. We need to rest at 1101 level of significance if the method employed does yield 95% confidence interval. (1) Ha. I p = [3.95 (zjnuzp at 5.95 (3)11 = 0.01 x 454 p.;] U.95 (4)2:n= L=_225 M (\"-953 9-55 3'1 50:] (5] Based one = 0.01. 2'53 = 30.005 = 2.5TII or 2.53. (5) Since |2| Zay. we reject Ho. (b) at 0.01 level of significance. (1) Ho : p = 0.30 (2) Ha:p = 0.30 (3) a = 0.01 157 - Po 600 (0.30) (4) z = -2.05 Po(1 - Po) (0.30) (0.70) 600 (5) Based on a = 0.01, Zay, = Zo.ous = 2.57 or 2.58 (6) Since |z|