Question

an old medical textbook states that the mean sodium level for healthy adults is 141 meq per liter of blood. a medical researcher believes that, because of modern dietary habits, the mean sodium level for healthy adults, μ, now differs from that given in the textbook. a random sample of 48 healthy adults is evaluated. the mean sodium level for the sample is 138 meq per liter of blood. it is known that the population standard deviation of adult sodium levels is 13 meq. assume that the population is normally distributed. can we conclude, at the 0.01 level of significance, that the population mean adult sodium level differs from that given in the textbook? perform a two - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places, and round your responses as specified below. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the two critical values. (round to three or more decimal places.) □ and □ (e) can we conclude that the population mean adult sodium level differs from that given in the textbook? yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the population mean $\mu$ is equal to the textbook - given mean. The alternative hypothesis $H_1$ is that the population mean $\mu$ is not equal to the textbook - given mean.
$H_0:\mu = 141$
$H_1:\mu
eq141$

Step2: Determine test - statistic type

Since the population standard deviation $\sigma$ is known ($\sigma = 13$), and the population is normally distributed, we use the z - test statistic. The formula for the z - test statistic is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean under the null hypothesis, $\sigma$ is the population standard deviation, and $n$ is the sample size. Here, $\bar{x}=138$, $\mu = 141$, $\sigma = 13$, and $n = 48$.

Step3: Calculate the test - statistic value

$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{138 - 141}{\frac{13}{\sqrt{48}}}\approx\frac{- 3}{\frac{13}{6.928}}\approx\frac{-3}{1.877}\approx - 1.609$

Step4: Find critical values

For a two - tailed test with a significance level of $\alpha=0.01$, the critical values are $z_{\alpha/2}$ and $-z_{\alpha/2}$. Since $\alpha = 0.01$, then $\alpha/2=0.005$. Looking up in the standard normal distribution table, $z_{0.005}\approx2.576$ and $-z_{0.005}\approx - 2.576$.

Step5: Make a decision

Since the calculated z - value ($z=-1.609$) is between the critical values ($-2.576$ and $2.576$), we fail to reject the null hypothesis.

Answer:

(a) $H_0:\mu = 141$; $H_1:\mu
eq141$
(b) z - test statistic
(c) $-1.609$
(d) $-2.576$ and $2.576$
(e) No