onsider the hypotheses shown below. Given that x overbarxequals=108108​, sigmaσequals=2525​, nequals=4242​, alphaαequals=0.010.01​, complete parts a through c below.
Upper H 0H0​: muμequals=115115
Upper H Subscript Upper AHA​: muμnot equals≠115115
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Part 1
a. State the decision rule in terms of the critical​ value(s) of the test statistic.
Reject the null hypothesis if the calculated value of the test​ statistic,
▼
t,
z,
is
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not contained within
greater than
equal to
less than
the critical​ value(s), enter your response here. ​Otherwise, do not reject the null hypothesis.

We are performing a hypothesis test for the population mean \(\mu\), with the null hypothesis \( H_0: \mu = 115 \) and the alternative hypothesis \( H_A: \mu \neq 115 \). This is a two-tailed test since the alternative hypothesis specifies that \(\mu\) is not equal to 115.

### Given information:
- Sample mean (\(\overline{x}\)) = 108
- Population standard deviation (\(\sigma\)) = 25
- Sample size (\(n\)) = 42
- Significance level (\(\alpha\)) = 0.01
- \( H_0: \mu = 115 \)
- \( H_A: \mu \neq 115 \)

### a. State the decision rule in terms of the critical values of the test statistic.

Since we know the population standard deviation, we will use a **z-test** (normal distribution) to calculate the test statistic. For a two-tailed z-test, the decision rule is based on the critical values of the standard normal distribution corresponding to the significance level \(\alpha\).

#### Step 1: Find the critical values
Since this is a two-tailed test, we split the significance level \(\alpha = 0.01\) into two equal parts (for the two tails), so each tail has a probability of \(\frac{\alpha}{2} = 0.005\).

From the z-table, the critical z-values for \(\alpha/2 = 0.005\) are:
- \( z_{\text{critical}} = \pm 2.576\)

#### Step 2: Decision rule
The decision rule is:
- Reject the null hypothesis if the calculated value of the test statistic (z) is **not contained within** the interval \([-2.576, 2.576]\).
- Otherwise, do not reject the null hypothesis.

### Summary:
- **Reject \(H_0\)** if \( z \) is **not contained within** the interval \([-2.576, 2.576]\).
- **Do not reject \(H_0\)** if \( z \) is contained within this interval.

Would you like to proceed with the next parts of the question? If yes, we can continue to calculate the test statistic and make a decision based on this rule.

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#### Follow-up questions:
1. What is the formula to compute the z-test statistic in a hypothesis test for the mean?
2. How would the decision rule change if this were a one-tailed test?
3. What are the implications of changing the significance level \(\alpha\) to 0.05 in a two-tailed test?
4. How does the sample size affect the width of the confidence interval in hypothesis testing?
5. Can you explain how to interpret a z-score in the context of this hypothesis test?

#### Tip:
Always check whether you're dealing with a one-tailed or two-tailed test, as it directly affects the decision rule and critical values.

Consider the hypotheses shown below. Given that x̄ = 108, σ = 25, n = 42, α = 0.01, complete parts a through c below. H₀: μ = 115, Hₐ: μ ≠ 115. Part 1: a. State the decision rule in terms of the critical value(s) of the test statistic. Reject the null hypothesis if the calculated value of the test statistic is not contained within the critical value(s). Otherwise, do not reject the null hypothesis.

Learn how to conduct a two-tailed z-test for the population mean with sample data, x̄ = 108, σ = 25, n = 42, and α = 0.01. This tutorial explains the critical value decision rule and walks through the step-by-step process of hypothesis testing. Suitable for undergraduate statistics students.

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