Hypothesis Testing for Population Mean (σ Unknown) – Step-by-Step Guidance

Q1. Use the P-value method with a 0.05 significance level to test the claim that the mean amount of sleep for adults is less than 7 hours.

Background

Topic: Hypothesis Testing for Population Mean (σ Unknown)

This question tests your ability to conduct a hypothesis test for a population mean when the population standard deviation is not known, using the t-distribution and the P-value method.

Key Terms and Formulas

• Null hypothesis ($H_0$): $\mu = 7$

• Alternative hypothesis ($H_a$): $\mu < 7$

• Sample mean ($\bar{x}$): 6.833...

• Sample standard deviation ($s$): 1.592...

• Sample size ($n$): 12

• Degrees of freedom ($df$): $n - 1$

• Test statistic formula: $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu = 7$, $H_a: \mu < 7$ (claim is that the mean is less than 7 hours).

2. Identify the sample statistics: $\bar{x} = 6.833...$, $s = 1.592...$, $n = 12$.

3. Calculate the degrees of freedom: $df = n - 1 = 11$.

4. Set up the test statistic formula: $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$.

5. Plug in the values for $\bar{x}$, $\mu$, $s$, and $n$ to compute the t-statistic, but stop before calculating the final value.

Try solving on your own before revealing the answer!

Final Answer: t = -0.36, P-value = 0.36 (Fail to reject $H_0$)

Using the formula $t = \frac{6.833 - 7}{1.592 / \sqrt{12}}$, the calculated t-statistic is approximately -0.36. The P-value is greater than 0.05, so we do not have enough evidence to support the claim that the mean is less than 7 hours.

Q2. Use a 0.05 significance level to test the common belief that the population mean body temperature is 98.6°F.

Background

Topic: Hypothesis Testing for Population Mean (σ Unknown)

This question tests your ability to use the t-distribution to test a claim about a population mean when the population standard deviation is not known, using sample statistics.

Key Terms and Formulas

• Null hypothesis ($H_0$): $\mu = 98.6$

• Alternative hypothesis ($H_a$): $\mu \neq 98.6$

• Sample mean ($\bar{x}$): 98.20

• Sample standard deviation ($s$): 0.62

• Sample size ($n$): 106

• Degrees of freedom ($df$): $n - 1$

• Test statistic formula: $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$

Step-by-Step Guidance

1. State the hypotheses: $H_0: \mu = 98.6$, $H_a: \mu \neq 98.6$ (two-tailed test).

2. Identify the sample statistics: $\bar{x} = 98.20$, $s = 0.62$, $n = 106$.

3. Calculate the degrees of freedom: $df = n - 1 = 105$.

4. Set up the test statistic formula: $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$.

5. Plug in the values for $\bar{x}$, $\mu$, $s$, and $n$ to compute the t-statistic, but stop before calculating the final value.

Try solving on your own before revealing the answer!

Final Answer: t = -6.13, P-value < 0.05 (Reject $H_0$)

Using the formula $t = \frac{98.20 - 98.6}{0.62 / \sqrt{106}}$, the calculated t-statistic is approximately -6.13. The P-value is much less than 0.05, so we reject the null hypothesis and conclude that the mean body temperature is significantly different from 98.6°F.

Q3. What is the critical value for a t-test with n = 12?

Background

Topic: t-Distribution Critical Values

This question tests your understanding of how to find the critical value for a t-test using the Student t-distribution and degrees of freedom.

Key Terms and Formulas

• Degrees of freedom ($df$): $n - 1$

• Critical value: The value from the t-distribution table corresponding to the desired significance level and degrees of freedom.

Step-by-Step Guidance

1. Calculate the degrees of freedom: $df = n - 1 = 11$.

2. Identify the significance level (e.g., 0.05 for a two-tailed test).

3. Look up the critical value in the t-distribution table for $df = 11$ and the specified significance level.

4. If the exact $df$ is not available, use the next lower $df$ or interpolate between values.

Try solving on your own before revealing the answer!

Final Answer: tcritical ≈ 2.201 (for df = 11, two-tailed, α = 0.05)

For $df = 11$ and a two-tailed test at the 0.05 significance level, the critical value is approximately 2.201. This value is used to determine the rejection region for the hypothesis test.