Questions: 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>1170 ; α=0.01 ; σ=214.43. Sample statistics: x̄=1190.46, n=250
Identify the null and alternative hypotheses. Choose the correct answer below.
A. H0: μ ≥ 1190.46 Ha: μ<1190.46
C. H0: μ ≤ 1190.46 Ha: μ>1190.46
E. H0: μ>1170 Ha: μ ≤ 1170
F. H0: μ>1190.46 Ha: μ ≤ 1190.46
Calculate the standardized test statistic, z. Use the z formula.
The standardized test statistic, z is . (Round to two decimal places as needed.)
Transcript text: 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: $\mu>1170 ; \alpha=0.01 ; \sigma=214.43$. Sample statistics: $\bar{x}=1190.46, n=250$
Identify the null and alternative hypotheses. Choose the correct answer below.
A. $H_{0}: \mu \geq 1190.46$ $H_{a}: \mu<1190.46$
C. $H_{0}: \mu \leq 1190.46$ $H_{a}: \mu>1190.46$
E. $H_{0}: \mu>1170$
$H_{a}: \mu \leq 1170$
F.
\[
\begin{array}{l}
\mathrm{H}_{0}: \mu>1190.46 \\
\mathrm{H}_{\mathrm{a}}: \mu \leq 1190.46
\end{array}
\]
Calculate the standardized test statistic, $z$. Use the $z$ formula.
The standardized test statistic, $z$ is $\square$ I.
(Round to two decimal places as needed.)
Solution
Solution Steps
Step 1: Identify the Hypotheses
The null and alternative hypotheses are defined as follows:
Null Hypothesis (\( H_0 \)): \( \mu \leq 1170 \)
Alternative Hypothesis (\( H_a \)): \( \mu > 1170 \)
Step 2: Calculate the Standardized Test Statistic
The standardized test statistic \( z \) is calculated using the formula:
\[
z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}
\]
Substituting the values:
\[
z = \frac{1190.46 - 1170}{\frac{214.43}{\sqrt{250}}}
\]
This results in:
\[
z \approx 1.51
\]
Step 3: Determine the P-value
To find the P-value for the right-tailed test, we calculate:
\[
P(Z > z) = 1 - P(Z \leq z)
\]
Using the cumulative distribution function (CDF) for the standard normal distribution, we find:
\[
P(Z \leq 1.51) \approx 0.9345
\]
Thus, the P-value is:
\[
P(Z > 1.51) \approx 1 - 0.9345 = 0.0655
\]
Final Answer
The null and alternative hypotheses are:
\( H_{0}: \mu \leq 1170 \) and \( H_{a}: \mu > 1170 \).
The standardized test statistic, \( z \), is \( \boxed{1.51} \).