Question

Test the hypothesis that the half-life of a bulb is 1600 hr. against the alternative that it is 1570 hr. For engineering reasons, only 16 bulbs could be tested obtaining an average of 1590 hr. with a standard deviation of 30 hr. Use a 99% confidence level.

A) Ho is not rejected. z = -1.33> -2.602

B) Ho is rejected. t = -1.33> -2.602

C) Ho is rejected. z = -1.33> -2.602

D) Ho is not rejected. t = -1.33> -2.602

Answer 1

Solution:

Since the population standard deviation is unknown , we use t test.

The null and alternative hypothesis are

H₀ : $\mu$ = 1600 vs H_a : $\mu$ <   1600

The test statistic t is

t =  
T-]/[s/n
= [1590 - 1600]/[30 /$\sqrt{}$16] = -1.33

Now ,

d.f. = n - 1 = 16- 1 = 15

c = 99% = 0.99

$\alpha$ = 1 - c = 1 - 0.99 = 0.01

< sign in Ha indicates that the test is LEFT TAILED.

So critical value is

-ta,df -0.01,15 = -2.602

Rejection region is < -2.602

But

-1.33 > -2.602

Therefore ,

D) Ho is not rejected. t = -1.33 > -2.602