sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the reguired hypothesis test about the mean, u, of the population from which the šample was drawn. ritical-value approach. = 40.4, s = 5.9, n= 15, Ho: H= 32.6, H:#32.6, a= 0.05. O A. Test statistic: t=5.12. Critical values: t= +1.96. Do not reject Ho: u= 32.6. There is not sufficient evidence to support the claim that the mean is different from 32.6. O B. Test statistic: t=5.12. Critical values: t= +2.145. Reject Ho: u= 32.6. There is sufficient evidence to support the claim that the mean is different from 32.6. O C. Test statistic: t= 5.12. Critical values: t= +1.96. Reject Ho: u= 32.6. There is sufficient evidence to support the claim that the mean is different from 32.6. O D. Test statistic: t= 5.12. Critical values: t= +2.145. Do not reject Ho: H= 32.6. There is not sufficient evidence to support the claim that the mean is different from 32.6.

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A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean, μ, of the population from which the sample was drawn. Use the critical-value approach.

Given:  
- \( \bar{x} = 40.4 \)
- \( s = 5.9 \)
- \( n = 15 \)
- \( H_0: \mu = 32.6 \)
- \( H_a: \mu \neq 32.6 \)
- \( \alpha = 0.05 \)

Options:

**A.**
- Test statistic: \( t = 5.12 \)
- Critical values: \( t = \pm 1.96 \)
- Conclusion: Do not reject \( H_0: \mu = 32.6 \). There is not sufficient evidence to support the claim that the mean is different from 32.6.

**B.**
- Test statistic: \( t = 5.12 \)
- Critical values: \( t = \pm 2.145 \)
- Conclusion: Reject \( H_0: \mu = 32.6 \). There is sufficient evidence to support the claim that the mean is different from 32.6.

**C.**
- Test statistic: \( t = 5.12 \)
- Critical values: \( t = \pm 1.96 \)
- Conclusion: Reject \( H_0: \mu = 32.6 \). There is sufficient evidence to support the claim that the mean is different from 32.6.

**D.**
- Test statistic: \( t = 5.12 \)
- Critical values: \( t = \pm 2.145 \)
- Conclusion: Do not reject \( H_0: \mu = 32.6 \). There is not sufficient evidence to support the claim that the mean is different from 32.6.
Transcribed Image Text:A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean, μ, of the population from which the sample was drawn. Use the critical-value approach. Given: - \( \bar{x} = 40.4 \) - \( s = 5.9 \) - \( n = 15 \) - \( H_0: \mu = 32.6 \) - \( H_a: \mu \neq 32.6 \) - \( \alpha = 0.05 \) Options: **A.** - Test statistic: \( t = 5.12 \) - Critical values: \( t = \pm 1.96 \) - Conclusion: Do not reject \( H_0: \mu = 32.6 \). There is not sufficient evidence to support the claim that the mean is different from 32.6. **B.** - Test statistic: \( t = 5.12 \) - Critical values: \( t = \pm 2.145 \) - Conclusion: Reject \( H_0: \mu = 32.6 \). There is sufficient evidence to support the claim that the mean is different from 32.6. **C.** - Test statistic: \( t = 5.12 \) - Critical values: \( t = \pm 1.96 \) - Conclusion: Reject \( H_0: \mu = 32.6 \). There is sufficient evidence to support the claim that the mean is different from 32.6. **D.** - Test statistic: \( t = 5.12 \) - Critical values: \( t = \pm 2.145 \) - Conclusion: Do not reject \( H_0: \mu = 32.6 \). There is not sufficient evidence to support the claim that the mean is different from 32.6.
Expert Solution
Step 1: Given information

According to the given information, we have

H0: μ=32.6H1: μ32.6

Sample mean = 40.4

Sample standard deviation = 5.9

Sample size = 15

Level of significance = 0.05

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