Question 1

The life in hours of a battery is known to be approximately normally distributed with standard deviation \(\sigma=1.25\) hours. A random sample of 10 batteries has a mean life of \(\bar{x}=40.5\) hours.

A. Is there evidence to support the claim that battery life exceeds 40 hours? Use \(\alpha=0.05\).

For the first question, let’s start off by listing down the null and alternative hypotheses, and all the given.

\(H_0:\mu\leq40\)

\(H_a:\mu>40\)

Given:

\(\sigma=1.25\) hours

\(n=10\) batteries

\(\bar{x}=40.5\) hours

\(\mu=40\) hours

\(\alpha=0.05\)

To test the claim, we will use Z-test to determine whether the two population means are different. We can use the formula given below and plug in the values above.

\[ Z_0=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \] \[ =\frac{40.5-40}{\frac{1.25}{\sqrt{10}}} \] \[ =1.264911064\approx1.26 \]

We will now find the \(P\)-value and compare it to the significance level \(\alpha=0.05\).

\[ P-value=1-\Phi(Z_0)=1-\Phi(1.26)=1-0.896165=0.103835 \]

Since \(0.103835>0.05\), we fail to reject the null hypothesis.

\(\therefore\) There is not enough evidence to say that the battery life exceeds 40 hours.

B. What is the \(P\)-value for the test in part A?

The \(P\)-value is the probability of rejecting the null hypothesis when it is true. In order to get this value, we just need to use the Standard Normal Cumulative Probability Table and find the corresponding value for \(Z_0\).

\[ P-value=1-\Phi(Z_0)=1-\Phi(1.26)=1-0.896165=0.103835 \]

We can also use code chunks in R to find this value.

p <- 1 - pnorm(q=1.26);
p

## [1] 0.1038347

\(\therefore\) The \(P\)-value is 0.103835.

C. What is the \(\beta\)-error for the text in part B if the true mean life is 42 hours?

Given:

\(\alpha=0.05\)

\(\delta=\mu-\mu_0=42-40=2\)

\(n=10\)

\(\sigma=1.25\)

\(\beta\)-error is the probability of failing to reject the null hypothesis when it is false. In order to get this value, we just need to use the formula below and substitute the given above.

\[ \beta=\Phi(z_\alpha-\frac{\delta\sqrt{n}}{\sigma})=\Phi(1.64-\frac{2\sqrt{10}}{1.25})=\Phi(-3.419644256)=\Phi(-3.42)=0.000313 \]

Again, we can also use code chunks in R to find its value.

beta <- pnorm(q=-3.42);
beta

## [1] 0.0003131057

\(\therefore\) The \(\beta\)-error is 0.000313.

D. What sample size would be required to ensure that \(\beta\) does not exceed 0.10 if the true mean is 44 hours?

Given:

\(\beta=0.10\)

\(\sigma=1.25\)

\(\delta=\mu-\mu_0=44-40=4\)

We do not want to make the probability of type II error occurring to exceed 0.10 or \(\beta<0.10\). In order to do that, we will use the formula below and substitute the following given.

\[ n=\frac{(Z_\alpha+Z_\beta)^2\sigma^2}{\delta^2}=\frac{(1.64+1.28)^2(1.25)^2}{4^2}=0.83265625\approx1 \]

\(\therefore\) A sample size of 1 is required to ensure that the \(\beta\)-error does not exceed 0.10.

E. Explain how you could answer the question in part A by calculating an appropriate confidence bound on battery life.

Given:

\(\bar{x}=40.5\)

\(\alpha=0.05\)

\(\sigma=1.25\)

\(n=10\)

With the given problem, the confidence interval should be one-sided. The formula below will be used and the given above will be substituted to find the confidence interval.

\[ \bar{x}+Z_\alpha(\frac{\sigma}{\sqrt{n}})=40.5+(1.64)(\frac{1.25}{\sqrt{10}})=41.14826692\approx41.15 \]

Our interval will then be \((-\infty,41.15)\). The battery life of 40 hours is inside the confidence interval. That means that we fail to reject the null hypothesis.

\(\therefore\) There is not enough evidence to reject the claim that the battery life exceeds 40 hours.