Question

1) The life in hours of a battery is known to be approximately normally distributed with standard deviation o = 1.25 hours. A random sample of 10 batteries has a sample mean life 40.5 hours. Is there evidence to support the claim that battery life exceeds 40 hours? Use a = 0.5. Zo = 1.26 < 1.65 fail to reject H. What is the P-value for the test in part (a)? P-value = 0.1038 Explain how you could answer the question in part (a) by calculating an appropriate confidence bound on battery life. 39.85 su

*Show all work, the answer is given already*

Answer 1

Solution:

The null and alternative hypothesis are

H₀ : $\mu$ = 40 vs H_a : $\mu$ > 40

The test statistic t is

z₀ = ($url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1596180112&ymreqid=f4764063-0d40-e2b0-1cde-f00115012f00&sig=YvAS7FjHffzTvoIlZIr2Zg--~D$ - $\mu$ )/[$\sigma$/$\sqrt{}$n] = [40.5- 40]/[1.25/$\sqrt{}$10] = 1.26

$\alpha$ = 0.05

Right tailed test

Critical value is

2 = 20.05 1.65

z₀ = 1.26 < 1.65 , so , fail to reject H₀

Now , for right tailed test ,

p value = P(Z > z₀] = P[Z > 1.26] = 1 - P[Z < 1.26] = 1 - 0.8962 = 0.1038

p value = 0.1038

Now , we construct one sided lower confidence interval

Margin of error = E =
Za *[/vn
= 1.65 * [1.25/$\sqrt{}$10] = 0.65

Lower bound = $url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1596180112&ymreqid=f4764063-0d40-e2b0-1cde-f00115012f00&sig=YvAS7FjHffzTvoIlZIr2Zg--~D$ - E = 40.5 - 0.65 = 39.85

So ,

39.85  $\leq$  $\mu$

Now , we construct one sided upper confidence interval

Lower bound = $url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1596180112&ymreqid=f4764063-0d40-e2b0-1cde-f00115012f00&sig=YvAS7FjHffzTvoIlZIr2Zg--~D$ + E = 40.5 + 0.65 = 41.15

So ,

$\mu$  $\leq$  41.15

Since , hypothetical value 40 is not greater than 41.15 , so

fail to reject H₀