Question

test the claim about the population mean \\( \mu \\) at the level of significance \\( \alpha \\). assume the population is normally distributed.
claim: \\( \mu = 1,560; \alpha = 0.01; \sigma = 82 \\)
sample statistics: \\( \overline { x } = 1,530, n = 35 \\)
a. reject \\( h _ { 0 } \\). there is enough evidence at the \\( 1 \\% \\) level of significance to reject the claim.
b. fail to reject \\( h _ { 0 } \\). there is not enough evidence at the \\( 1 \\% \\) level of significance to reject the claim.
c. there is not enough information to decide.

Explanation:

Step1: State the hypotheses

The null hypothesis \(H_0:\mu = 1560\) (the claim), and the alternative hypothesis \(H_1:\mu
eq1560\) (two - tailed test).

Step2: Calculate the test statistic

The formula for the \(z\) - test statistic is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
Substitute \(\bar{x} = 1530\), \(\mu = 1560\), \(\sigma = 82\), and \(n = 35\) into the formula:

$$
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Step3: Find the critical values

For a two - tailed test with \(\alpha=0.01\), the critical values are \(z_{\alpha/2}=\pm z_{0.005}\). From the standard normal table, \(z_{0.005}=\pm2.576\)

Step4: Make a decision

Since \(-2.576<-2.16 < 2.576\) (the test statistic \(z=-2.16\) does not fall in the rejection region), we fail to reject the null hypothesis \(H_0\).

Answer:

B. Fail to reject \(H_0\). There is not enough evidence at the \(1\%\) level of significance to reject the claim.