Question

a fast food restaurant estimates that the mean sodium content in one of its breakfast sandwiches is no more than 927 milligrams. a random sample of 39 breakfast sandwiches has a mean sodium content of 919 milligrams. assume the population standard deviation is 16 milligrams. at α = 0.10, do you have enough evidence to reject the restaurants claim? complete parts (a) through (e).
(b) identify the critical value(s). use technology.

z₀ = 1.28
(use a comma to separate answers as needed. round to two decimal places as needed.)

identify the rejection region(s). select the correct choice below.

a. the rejection regions are z > 1.28 and z < - 1.28.
b. the rejection region is z < 1.28.
c. the rejection region is z > 1.28.

Explanation:

Step1: Analyze the claim

The restaurant's claim is $\mu\leq927$. This is a left - tailed test.

Step2: Recall the rule for critical value and rejection region in left - tailed test

For a left - tailed test with significance level $\alpha = 0.10$, the critical value $z_0$ is such that $P(Z 927$ (if we use the formula $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, when we want to reject the null hypothesis $\mu\leq927$, we are looking for large positive values of $z$). Wait, no, actually, the claim is $\mu\leq927$ (null hypothesis $H_0:\mu\leq927$ and alternative hypothesis $H_1:\mu > 927$). For a right - tailed test (since $H_1:\mu > 927$), the critical value $z_0$ is such that $P(Z > z_0)=\alpha$. Using technology (e.g., in Excel: =NORM.S.INV(1 - 0.10)), we get $z_0 = 1.28$. The rejection region for a right - tailed test is $z>z_0$.

Answer:

C. The rejection region is $z > 1.28$