Consider the following hypothesis statement usingα=0.01and data from two independent samples. Assume the population variances are equal and the populations are normally distributed. Complete parts a and b.H0: μ1−μ2=0x1=14.5x2=13.0H1: μ1−μ2≠0s1=2.6s2=3.3 n1=22n2=15 a. Calculate the appropriate test statistic and interpret the result.The test statistic is?The critical value(s) is(are)?b. Identify the p-value from part a and interpret the result.

Introduction: Denote μ1, μ2 as the true population means. The null and alternative hypotheses are as follows: H0: μ1 = μ2 versus H1: μ1 ≠ μ2.a. Since the population standard deviations are unknown and the samples are taken from two independent populations, the independent samples t-test is to be conducted. The respective sample sizes are: n1 = 22 and n2 = 15. The respective sample means are: x̅1 = 14.5 and x̅2 = 13.0. The respective sample standard deviations are: s1 = 2.6 and s2 = 3.3. The degrees of freedom for the test is, df = n1 + n2 – 2 = 36 + 36 – 2 = 70. The pooled variance is: sp2 = [(n1 – 1) s12 + (n2 – 1) s22] / (n1 + n2 – 2) = [(21 × (2.6)2) + (14 × (3.3)2)] / 35 = 8.412. The test statistic is found as follows: t = (x̅1 – x̅2) / √ [sp2 (1/n1 + 1/n2)] = (14.5 – 13.0) / √ [(8.412) ∙ (1/22 + 1/15)] ≈ (1.5) / (0.9712) ≈ 1.545. Thus, the independent samples t-test statistic is 1.545. The level of significance is, α = 0.01. The critical values for the two-tailed test at 100 (1 – α) % level of confidence, that is, level of significance α are ± tα/2; n–2 when there are n pairs of observations, such that, PHo (| t n1 + n2 – 2 | ≥ | tα/2; n1 + n2 – 2 |) = α. Thus, the critical values here are ± t0.005;35, such that P (| t35 | ≥ | t0.005;35 |) = 0.01. Using the Excel formula: =T.INV.2T(0.01,35), P (| t35 | ≥ | 2.724 |) ≈ 0.01. Hence, the critical values are, ± 2.724.b. Since –2.724 < observed t (= 1.545) < 2.724, the p-value must be greater than 0.01. The actual p-value is calculated below: The p-value is, p-value = P (| t35 | ≥ t) = P (| t35 | ≥ | 1.545 |). Using the Excel formula: =T.DIST.2T(1.545,35), P (| t35 | ≥ | 1.545 |) ≈ 0.1313. Thus, the p-value is approximately 0.1313. Decision rule using p-value: Reject H0 at significance level α, if p-value ≤ α. Otherwise, fail to reject H0. Since p-value (≈ 0.1313) > α (= 0.01), fail to reject the null hypothesis, H0. Hence, the conclusion at α = 0.01 is that, there is not enough evidence to suggest that the populations means are significantly different.

Answered: Consider the following hypothesis…

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Consider the following hypothesis statement using

α=0.01

and data from two independent samples. Assume the population variances are equal and the populations are normally distributed. Complete parts a and b.

H0: μ1−μ2=0	x1=14.5	x2=13.0
H1: μ1−μ2≠0	s1=2.6	s2=3.3
	n1=22	n2=15

a. Calculate the appropriate test statistic and interpret the result.

The test statistic is?

The critical value(s) is(are)?

b. Identify the p-value from part a and interpret the result.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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