Question

Problem 6 Calculate the test statistic. (A z-score and a t-score are examples of test statistics.) In order to present your final results with a numerical accuracy of 0.01, it will be necessary to perform your intermediate calculations with an accuracy of 0.001. Part 1 The cost of a daily newspaper varies from city to city. A study was done to test the claim that the average cost of a daily newspaper is $0.35 . Twenty cities were surveyed; the average cost was $0.30 with a standard deviation of $0.03 . t= 7.45 Part 2 Your statistics instructor claims that 60 percent of the students who take an Elementary Statistics class go through life feeling more enriched. For some mysterious reason, most people don’t believe this. You decide to check this out on your own. You randomly survey 80 former Elementary Statistics students and find that 25 feel more enriched as a result of the class. z= 2.90

Answer 1

Answer:

Check the explanation

Step-by-step explanation:

1)  The null and alternative hypothesis is ,

$H_0 : \mu = 0.35$

$H_a : \mu \neq 0.35$

Test statistic = t =

= $(T - \mu ) / s / \sqrt n$

= $(0.30 - 0.35 ) / 0.03 / \sqrt 20$

Test statistic = t = - 7.45

2) Solution :

The null and alternative hypothesis is

$H_0$ : p = 0.60

$H_a$ : p > 0.60

$\hat p$ = x / n = 25 / 60 = 0.417

P0 = 0.60

1 - P0 = 1 - 0.60 = 0.40

Test statistic = z = 0.417

= $\hat p - P0 / [\sqrtP0 * (1 - P0 ) / n]$

= $0.417 - 0.60/ [\sqrt(0.60 *(1 - 0.60) / 80 ]$

Test statistic = z = -3.34