Question

question 30 of 33 stacked a confidence interval for the population mean \\( \mu \\) tells us which values of \\( \mu \\) are plausible (those inside the interval) and which values are not plausible (those outside the interval) at the chosen level of confidence. you can use this idea to carry out a test of any null hypothesis \\( h_0: \mu = \mu_0 \\) starting with a confidence interval: reject \\( h_0 \\) if \\( \mu_0 \\) is outside the interval and fail to reject if \\( \mu_0 \\) is inside the interval. the alternative hypothesis is always two - sided, \\( h_a: \mu \
eq \mu_0 \\) because the confidence interval extends in both directions from \\( \bar{x} \\). a 95% confidence interval leads to a test at the 5% significance level because the interval is wrong 5% of the time. in general, a confidence level \\( c \\) leads to a test at significance level \\( \alpha = 1 - c \\). a medical director found the ldl cholesterol \\( \bar{x} = 124.86 \\) for an srs of 72 executives. the standard deviation of the ldl of all adults is \\( \sigma = 40 \\). find the p - value for the z test. give your answer to four decimal places. p - value: macmillan learning

Explanation:

Step1: Identify the test type and parameters

We have a z - test for a population mean. The sample mean \(\bar{x}=124.86\), sample size \(n = 72\), population standard deviation \(\sigma=40\). We assume the null hypothesis \(H_0:\mu=\mu_0\) (but since we are doing a z - test for the mean, we can calculate the z - statistic first. Wait, actually, maybe we missed the value of \(\mu_0\)? Wait, no, maybe in the context of the confidence interval and hypothesis testing, but the problem is to find the P - value for the z - test. Wait, perhaps the null hypothesis is that \(\mu = 130\) (a common value for LDL, maybe a typo or missing information? Wait, no, let's re - read. Wait, the problem says "A medical director found the LDL cholesterol \(\bar{x}=124.86\) for an SRS of 72 executives. The standard deviation of the LDL of all adults is \(\sigma = 40\)". Maybe we are testing \(H_0:\mu=130\) (a typical value). Let's proceed with that assumption (since otherwise we can't calculate).

The formula for the z - statistic in a one - sample z - test is \(z=\frac{\bar{x}-\mu_0}{\frac{\sigma}{\sqrt{n}}}\)

Let's assume \(\mu_0 = 130\) (a common target for LDL). Then:

\(z=\frac{124.86 - 130}{\frac{40}{\sqrt{72}}}\)

Step2: Calculate the z - statistic

First, calculate the standard error \(SE=\frac{\sigma}{\sqrt{n}}=\frac{40}{\sqrt{72}}\approx\frac{40}{8.4853}\approx4.714\)

Then, the numerator: \(124.86−130=- 5.14\)

So, \(z=\frac{-5.14}{4.714}\approx - 1.09\)

Step3: Calculate the P - value

Since the alternative hypothesis is two - tailed (because the confidence interval is two - sided, as per the text "The alternative hypothesis is always two - sided, \(H_a:\mu
eq\mu_0\)"), the P - value is \(2\times P(Z < -|z|)\)

We have \(z\approx - 1.09\), so \(|z| = 1.09\)

Looking up in the standard normal table, \(P(Z < - 1.09)=0.1379\)

Then the two - tailed P - value is \(2\times0.1379 = 0.2758\)

Wait, but maybe the \(\mu_0\) is different. Wait, maybe I made a wrong assumption. Wait, the text says "A confidence interval for the population mean \(\mu\) tells us...", but the problem is to find the P - value for the z - test. Wait, perhaps the null hypothesis is \(\mu = 130\). Let's check the calculation again.

Wait, \(\bar{x}=124.86\), \(\mu_0 = 130\), \(\sigma = 40\), \(n = 72\)

\(z=\frac{124.86 - 130}{\frac{40}{\sqrt{72}}}=\frac{-5.14}{\frac{40}{8.485281374}}\approx\frac{-5.14}{4.714045207}\approx - 1.09\)

The P - value for a two - tailed test is \(2\times\Phi(-|z|)\), where \(\Phi\) is the cumulative distribution function of the standard normal distribution.

\(\Phi(- 1.09)=0.1379\) (from standard normal table: the z - score of - 1.09 has a cumulative probability of 0.1379)

So, P - value \(=2\times0.1379 = 0.2758\)

Answer:

0.2758 (assuming \(\mu_0 = 130\), if \(\mu_0\) is different, the P - value will change. If there was a different \(\mu_0\) provided, the calculation would be adjusted accordingly)