A random sample from normal population yielded sample mean=40.8 and sample standard deviation= 6.1, n =...

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A random sample from normal population yielded sample mean=40.8 and sample standard deviation= 6.1, n =...

A random sample from normal population yielded sample mean=40.8 and sample standard deviation= 6.1, n = 15. H0: μ = 32.6, Ha: μ ≠ 32.6, α = 0.05. Perform the hypothesis test and draw your conclusion.

Question 3 options:

	Test statistic: t = 5.21. P-value=0.00013. Reject H0. There is sufficient evidence to support the claim that the mean is different from 32.6.
	Test statistic: t = 5.21. P-value=1.9E-7 (0.00000019). Do not reject H0. There is not sufficient evidence to support the claim that the mean is different from 32.6.
	Test statistic: t = 5.21. P-value=1.9E-7 (0.00000019). Reject H0. There is sufficient evidence to support the claim that the mean is different from 32.6.
	Test statistic: t = 5.21. P-value=0.00013. Do not reject H0. There is not sufficient evidence to support the claim that the mean is different from 32.6.

math Statistics-And-Probability

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Answer 1

Answer #1

Solution:

Using statistical software output of this problem is,

One sample T summary hypothesis test:

μ : Mean of population
H₀ : μ = 32.6
H_A : μ ≠ 32.6

Hypothesis test results:

Mean	Sample Mean	Std. Err.	DF	T-Stat	P-value
μ	40.8	1.5750132	14	5.2063055	0.0001

Therefore, P-value < 0.05, it is concluded that reject the null hypothesis.

The answer is,

Test statistic: t = 5.21. P-value=0.00013. Reject H0. There is sufficient evidence to support the claim that the mean is different from 32.6.

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Answer 2

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