Question

Suppose the general manager for a company that produces a complicated electronics component used in the navigation systems in automobiles has stated that 73% of the items made pass through the final inspection without needing any rework. The quality manager wished to verify this statement. To do so, she selected a random sample of n=140 parts at the final inspection station and found that 96 passed and did not require rework. a. State the appropriate null and alternative hypotheses for testing the general manager's claim. b. Using a = 0.10, from the sample data, what conclusion should the quality manager reach about the general manager's claim? Discuss. HA: p>0.73 b. Perform the test using a normal approximation. Identify the test statistic. - 1.18 (Round to two decimal places as needed.) Identify the p-value. (Round to three decimal places as needed.)

Answer 1

Solution:

a)

The null and alternative hypothesis are

H₀ : p = 0.73

H_A : p $\neq$ 0.73

b)

n = 140

x = 96

Let  $\hat p$ be the sample proportion.

$\hat p$ = x/n = 96/140 = 0.6857

The test statistic z is

z =   

PP p(1-p) n

=  (0.6857 - 0.73)/$\sqrt{}$[0.73*(1 - 0.73)/140]

= -1.18

Test statistic is

-1.18

Now , we find p value .

$\neq$ sign in Ha indicates that the test is TWO TAILED

For TWO TAILED test ,

p value = P(Z < -z) + P(Z > +z) = P(Z < -1.18) + P(Z > +1.18) = 0.1190+0.1190= 0.238

p value is

0.238

{

Fail to reject H₀

There is no evidence to reject the manager's claim

}