Question

use technology to help you test the claim about the population mean, μ, at the given level of significance, α, using the given sample statistics. assume the population is normally distributed.
claim: μ ≤ 1160; α = 0.08; σ = 199.88. sample statistics: (\bar{x}=1185.66), n = 200
identify the null and alternative hypotheses. choose the correct answer below.
a. (h_0: mu geq 1160) (h_a: mu < 1160)
b. (h_0: mu > 1160) (h_a: mu leq 1160)
c. (h_0: mu geq 1185.66) (h_a: mu < 1185.66)
d. (h_0: mu leq 1185.66) (h_a: mu > 1185.66)
e. (h_0: mu > 1185.66) (h_a: mu leq 1185.66)
f. (h_0: mu leq 1160) (h_a: mu > 1160)
calculate the standardized test statistic.
the standardized test statistic is (square).
(round to two decimal places as needed.)

Explanation:

Step1: Determine the hypotheses

The claim is $\mu\leq1160$. The null hypothesis $H_0$ is the statement of equality, and the alternative hypothesis $H_a$ is the statement we are trying to find evidence for. So $H_0:\mu\leq1160$ and $H_a:\mu > 1160$ (Option F is correct for hypotheses).

Step2: Calculate the standardized test statistic (z - score)

The formula for the z - score in a one - sample z - test is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$.
We are given $\bar{x} = 1185.66$, $\mu = 1160$, $\sigma=199.88$, and $n = 200$.
First, calculate $\frac{\sigma}{\sqrt{n}}=\frac{199.88}{\sqrt{200}}\approx\frac{199.88}{14.1421}\approx14.13$.
Then, $z=\frac{1185.66 - 1160}{14.13}=\frac{25.66}{14.13}\approx1.82$.

Answer:

The standardized test statistic is \(1.82\).