Hypothesis Test for Difference in Population Means ( $σ$ Unknown)
You wish to test the following claim $(H_{a})$ at a significance level of $α = 0.05$ .
$\begin{array}{l} H_{o} : μ_{1} = μ_{2} \\ H_{a} : μ_{1} \neq μ_{2} \end{array}$
You believe both populations are normally distributed, but you do not know the standard deviations for either. We will assume that the population variances are not equal.

You obtain a sample of size $n_{1} = 22$ with a mean of $M_{1} = 76.8$ and a standard deviation of $S D_{1} = 5$ . from the first population. You obtain a sample of size $n_{2} = 25$ with a mean of $M_{2} = 74.4$ and a standar deviation of $S D_{2} = 18.7$ from the second population.
What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic $=$
What is the $p$ -value for this sample? For this calculation, use the conservative under-estimate for the degrees of freedom. The degrees of freedom is the minimum of $n_{1} - 1$ and $n_{2} - 1$ : (Report answer accurate to four decimal places.)
$p$ -value $=$
The $p$ -value is...
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3. Find the p-value: Using a T-distribution table or calculator with T-score = 0.472 and DF = 21, find the p-value (two-tailed) = 0.6416

Steps

Step 1 ：1. Calculate the T-score: $t = \frac{(M_{1} - M_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\frac{S D_{1}^{2}}{n_{1}} + \frac{S D_{2}^{2}}{n_{2}}}$ = \frac{(76.8 - 74.4) - 0}{\sqrt{\frac{5^2}{22} + \frac{18.7^2}{25}}}\ = 0.472\)

Step 2 ：2. Calculate the degrees of freedom (DF): $D F = min (n_{1} - 1, n_{2} - 1) = min (21, 24) = 21$

Step 3 ：3. Find the p-value: Using a T-distribution table or calculator with T-score = 0.472 and DF = 21, find the p-value (two-tailed) = 0.6416

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