Question

The life in hours of a battery is known to be approximately normally distributed with standard deviation σ = 1.25 hours. A random sample of 10 batteries has a mean life of x =40.5hours. a. Is there evidence to support the claim that battery life exceeds 40 hours? Use α = 0.05. b. What is the P-value for the test in part (a)? c. What is the β-error for the test in part (a) if the true mean life is 42 hours? d. What sample size would be required to ensure that β does not exceed 0.10 if the true mean life is 44 hours? e. Explain how you could answer the question in part (a) by calculating an appropriate confidence bound on life.

Answer 1

Answer:

a) $z= \frac{40.5-40}{\frac{1.25}{\sqrt{10}}}=1.265$

Now we can calculate the p value since we have a right tailed test the p values is given by:

$p_v = P(Z>1.265) = 1-P(Z$

And since the $p_v >\alpha$ we have enough evidence to FAIL to reject the null hypothesis. So then there is not evidence to support the claim that the mean is greater than 40.

b) $p_v = P(Z>1.265) = 1-P(Z$

c) $\beta = P(Z< 1.645 - \frac{2\sqrt{10}}{1.25}) = P(Z$

d) We want to ensure that the probability of error type II not exceeds 0.1, and for this case we can use the following formula:

$n = \frac{(z_{\alpha} +z_{\beta})^2 \sigma^2}{(x-\mu)^2}$

The true mean for this case is $\mu = 44$ and we want $\beta$ so then $z_{1-0.1} =z_{0.9}=1.29$ represent the value on the normal standard distribution that accumulates 0.1 of the area on the right tail. And we can replace like this:

$n = \frac{(1.65+1.29)^2 1.25^2}{(44-40)^2} =0.844 \approx 1$

e) For this case we can calculate a one sided confidence interval given by:

$(-\infty , \bar x +z_{\alpha} \frac{\sigma}{\sqrt{n}})$

And if we replace we got:

$40.5 +1.65 \frac{1.25}{\sqrt{10}}=41.152$

And the confidence interval would be $(-\infty,41.152)$

And since 40 is on the confidence interval we don't have enough evidence to reject the null hypothesis on this case.

Step-by-step explanation:

Part a

We have the following data given:

$n =10$ represent the sample size

$\bar x = 40.5$ represent the sample mean

$\sigma =1.25$ represent the population deviation.

We want to test the following hypothesis:

Null: $\mu \leq 40$

Alternative:$\mu >40$

The significance level provided was $\alpha =0.05$

The statistic for this case since we have the population deviation is given by:

$z= \frac{\bar x -\mu}{\frac{\sigma}{\sqrt{n}}}$

If we replace the values given we got:

$z= \frac{40.5-40}{\frac{1.25}{\sqrt{10}}}=1.265$

Now we can calculate the p value since we have a right tailed test the p values is given by:

$p_v = P(Z>1.265) = 1-P(Z$

And since the $p_v >\alpha$ we have enough evidence to FAIL to reject the null hypothesis. So then there is not evidence to support the claim that the mean is greater than 40.

Part b

The p value on this case is given by:

$p_v = P(Z>1.265) = 1-P(Z$

Part c

For this case the probability of type II error is defined as the probability of incorrectly retaining the null hypothesis and is defined like this:

$\beta = P(Z< z_{\alpha} - \frac{(x-\mu)\sqrt{n}}{\sigma}})$

Where $z_{\alpha}=1.645$ represent the critical value for the test that accumulates 0.05 of the area on the right tail of the normal standard distribution.

The true mean on this case is assumed $\mu = 42$, so then we can replace like this:

$\beta = P(Z< 1.645 - \frac{2\sqrt{10}}{1.25}) = P(Z$

Part d

We want to ensure that the probability of error type II not exceeds 0.1, and for this case we can use the following formula:

$n = \frac{(z_{\alpha} +z_{\beta})^2 \sigma^2}{(x-\mu)^2}$

The ture mean for this case is $\mu = 44$ and we want $\beta$ so then $z_{1-0.1} =z_{0.9}=1.29$ represent the value on the normal standard distribution that accumulates 0.1 of the area on the right tail. And we can replace like this:

$n = \frac{(1.65+1.29)^2 1.25^2}{(44-40)^2} =0.844 \approx 1$

Part e

For this case we can calculate a one sided confidence interval given by:

$(-\infty , \bar x +z_{\alpha} \frac{\sigma}{\sqrt{n}})$

And if we replace we got:

$40.5 +1.65 \frac{1.25}{\sqrt{10}}=41.152$

And the confidence interval would be $(-\infty,41.152)$

And since 40 is on the confidence interval we don't have enough evidence to reject the null hypothesis on this case.