Question

use technology to help you test the claim about the population mean, \\( \mu \\), at the given level of significance, \\( \alpha \\), using the given sample statistics. assume the population is normally distributed.

claim: \\( \mu \leq 1160 ; \alpha=0.08 ; \sigma=199.88 \\). sample statistics: \\( \bar{x}=1185.66, n=200 \\)

identify the null and alternative hypotheses. choose the correct answer below.

\\( \bigcirc \\) a. \\( h_{0}: \mu \geq 1160 \\)
\\( h_{a}: \mu<1160 \\)
\\( \bigcirc \\) b. \\( h_{0}: \mu>1160 \\)
\\( h_{a}: \mu \leq 1160 \\)
\\( \bigcirc \\) c. \\( h_{0}: \mu \geq 1185.66 \\)
\\( h_{a}: \mu<1185.66 \\)
\\( \bigcirc \\) d. \\( h_{0}: \mu \leq 1185.66 \\)
\\( h_{a}: \mu>1185.66 \\)
\\( \bigcirc \\) e. \\( h_{0}: \mu>1185.66 \\)
\\( h_{a}: \mu \leq 1185.66 \\)
\\( \bigcirc \\) f. \\( h_{0}: \mu \leq 1160 \\)
\\( h_{a}: \mu>1160 \\)

calculate the standardized test statistic.

the standardized test statistic is 1.82
(round to two decimal places as needed.)

determine the p - value.

\\( p= \\) (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for the P - value in a right - tailed z - test

For a right - tailed z - test, \(P = 1-\Phi(z)\), where \(z\) is the standardized test statistic and \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.

Step2: Use the value of \(z\)

We are given \(z = 1.82\). Using a standard normal table or technology (e.g., a calculator with a normalcdf function: normalcdf\((1.82,\infty)\) or using the property \(P(Z>z)=1 - P(Z\leq z)\)).
From the standard normal table, \(P(Z\leq1.82)=0.9656\)

Step3: Calculate the P - value

\(P=1 - 0.9656=0.0344\approx0.034\)

Answer:

\(0.034\)