SOLUTION: A random sample of 10 observations was drawn from a large normally distributed population. The data is below. 14 12 20 21 18 22 18 14 12 17 Test to determine if we can infer

Question 1180121: A random sample of 10 observations was drawn from a large normally distributed population. The data is below.
14 12 20 21 18 22 18 14 12 17
Test to determine if we can infer at the 3% significance level that the population mean is not equal to 17, filling in the requested information below.
A. The value of the standardized test statistic:
Note: For the next part, your answer should use interval notation. An answer of the form (−∞,𝑎)
is expressed (-infty, a), an answer of the form (𝑏,∞) is expressed (b, infty), and an answer of the form (−∞,𝑎)∪(𝑏,∞) is expressed (-infty, a)U(b, infty).
B. The rejection region for the standardized test statistic:
C. Your decision for the hypothesis test:
A. Reject 𝐻0
B. Do Not Reject 𝐻0
C. Reject 𝐻1
D. Do Not Reject 𝐻1
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to conduct the hypothesis test:
**1. State the Hypotheses:**
* **Null Hypothesis (H0):** The population mean (μ) is equal to 17. (μ = 17)
* **Alternative Hypothesis (H1):** The population mean (μ) is *not* equal to 17. (μ ≠ 17) This is a two-tailed test.
**2. Significance Level:** α = 0.03
**3. Calculate the Sample Statistics:**
* **Sample Size (n):** 10
* **Sample Mean (x̄):**
x̄ = (14 + 12 + 20 + 21 + 18 + 22 + 18 + 14 + 12 + 17) / 10 = 168 / 10 = 16.8
* **Sample Standard Deviation (s):**
1. Calculate the squared deviations from the mean:
(14-16.8)^2 = 7.84
(12-16.8)^2 = 23.04
(20-16.8)^2 = 10.24
(21-16.8)^2 = 17.64
(18-16.8)^2 = 0.04
(22-16.8)^2 = 27.04
(18-16.8)^2 = 0.04
(14-16.8)^2 = 7.84
(12-16.8)^2 = 23.04
(17-16.8)^2 = 0.04
Sum of squared deviations = 116.8
s = √[Σ(xᵢ - x̄)² / (n - 1)] = √(116.8 / 9) ≈ √12.98 ≈ 3.603
**4. Calculate the Test Statistic (t-score):**
Since the sample size is small (n < 30) and the population standard deviation is unknown, we use a t-test.
t = (x̄ - μ) / (s / √n)
t = (16.8 - 17) / (3.603 / √10)
t = -0.2 / (3.603 / 3.162)
t = -0.2 / 1.139
t ≈ -0.176
**A. The value of the standardized test statistic:** -0.176
**5. Determine the Degrees of Freedom:**
Degrees of freedom (df) = n - 1 = 10 - 1 = 9
**6. Find the Critical t-values:**
For a two-tailed test with α = 0.03 and df = 9, we need to find the t-values that correspond to α/2 = 0.015 in each tail. Using a t-table or calculator, we find the critical t-values are approximately ±2.821.
**B. The rejection region for the standardized test statistic:** (-infty, -2.821)U(2.821, infty)
**7. Make a Decision:**
Compare the calculated t-statistic (-0.176) to the critical t-values (±2.821).
* -2.821 < -0.176 < 2.821
Since the calculated t-statistic falls *within* the range of the critical t-values, we *fail to reject* the null hypothesis.
**C. Your decision for the hypothesis test:** B. Do Not Reject H0

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