Problem

Next question You can retry this question below You wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.001$. \[ \begin{array}{l} H_{o}: \mu_{1}=\mu_{2} \\ H_{a}: \mu_{1}<\mu_{2} \end{array} \] You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal. You obtain a sample of size $n_{1}=17$ with a mean of $M_{1}=53.3$ and a standard deviation of $S D_{1}=9.7$ from the first population. You obtain a sample of size $n_{2}=26$ with a mean of $M_{2}=54.7$ and a standard deviation of $S D_{2}=5.5$ from the second population. What is the critical value for this test? For this calculation, use the conservative under-estimate for the degrees of freedom as mentioned in the textbook. (Report answer accurate to three decimal places.) critical value $=$ What is the test statistic for this sample? (Report answer, accurate to three decimal places.) test statistic $=$ The test statistic is...

Solution

Step 1 :Given the following values: \(n_1 = 17\), \(M_1 = 53.3\), \(SD_1 = 9.7\), \(n_2 = 26\), \(M_2 = 54.7\), \(SD_2 = 5.5\), and \(\alpha = 0.001\).

Step 2 :The degrees of freedom can be calculated as the smaller of \(n_1-1\) and \(n_2-1\), which gives us \(df = 16\).

Step 3 :Using the t-distribution table with the degrees of freedom and the significance level, we find the critical value to be approximately \(3.686\).

Step 4 :The test statistic can be calculated using the formula for the t-test statistic for two independent samples: \(t = \frac{M_1 - M_2}{\sqrt{\frac{SD_1^2}{n_1} + \frac{SD_2^2}{n_2}}}\).

Step 5 :Substituting the given values into the formula, we find the test statistic to be approximately \(-0.541\).

Step 6 :Final Answer: The critical value for this test is \(\boxed{3.686}\) and the test statistic for this sample is \(\boxed{-0.541}\).

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