## Related questions with answers

The null and alternate hypotheses are:

$H_0:μ_1≤μ_2; H_1:μ_1>μ_2$

A random sample of 20 items from the first population showed a mean of 100 and a standard deviation of 15. A sample of 16 items for the second population showed a mean of 94 and a standard deviation of 8. Use the .05 significant level.

Solution

VerifiedGiven:

$\begin{align*} n_1&=\text{Sample size}=20 \\ n_2&=\text{Sample size}=16 \\ \overline{x}_1&=\text{Sample mean}=100 \\ \overline{x}_2&=\text{Sample mean}=94 \\ s_1&=\text{Sample standard deviation}=15 \\ s_2&=\text{Sample standard deviation}=8 \\ \alpha&=\text{Significance level}=0.05 \\ H_0&:\mu_1\leq \mu_2 \\ H_1&:\mu_1>\mu_2 \end{align*}$

(a) Determine the degrees of freedom (rounded down to the nearest integer):

$\Delta=\dfrac{\left( \dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}\right)^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}}=\dfrac{\left( \dfrac{15^2}{20}+\dfrac{8^2}{16}\right)^2}{\dfrac{(15^2/20)^2}{20-1}+\dfrac{(8^2/16)^2}{16-1}}\approx 30$

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