Question

a school administrator wants to see if there is a difference in the number of students per class for portland public school district (group 1) compared to the beaverton school district (group 2). assume the populations are normally distributed with unequal variances. a random sample of 27 portland classes found a mean of 34 students per class with a standard deviation of 2. a random sample of 26 beaverton classes found a mean of 32 students per class with a standard deviation of 5. find a 95% confidence interval in the difference of the means. round answers to 4 decimal places. <\\( \mu _ { 1 } - \mu _ { 2 } < \\) enter an integer or decimal number more..

Explanation:

Step1: Calculate the difference in sample means

The difference in sample means $\bar{x}_1-\bar{x}_2$ is $34 - 32=2$.

Step2: Calculate the degrees of freedom

Using the formula $df=\frac{(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2})^{2}}{\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}+\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}}$.
Here, $s_1 = 2$, $n_1=27$, $s_2 = 5$, $n_2=26$.
$\frac{s_1^{2}}{n_1}=\frac{4}{27}\approx0.1481$, $\frac{s_2^{2}}{n_2}=\frac{25}{26}\approx0.9615$.
$\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}=0.1481 + 0.9615=1.1096$.
$(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2})^{2}=(1.1096)^{2}\approx1.2313$.
$\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}=\frac{(0.1481)^{2}}{26}\approx\frac{0.0219}{26}\approx0.00084$.
$\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}=\frac{(0.9615)^{2}}{25}\approx\frac{0.9245}{25}\approx0.03698$.
$\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}+\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}=0.00084+0.03698 = 0.03782$.
$df=\frac{1.2313}{0.03782}\approx32.56$. Using the conservative approach, we can round $df$ down to $25$ (or use a more precise method with technology). For a 95% confidence interval, the critical value $t_{\alpha/2}$ with $df = 25$ (using $t$-distribution table) is $t_{0.025,25}=2.060$.

Step3: Calculate the margin of error

The margin of error $E=t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}$.
Substitute $t_{\alpha/2}=2.060$, $\frac{s_1^{2}}{n_1}=0.1481$, $\frac{s_2^{2}}{n_2}=0.9615$.
$E = 2.060\sqrt{0.1481+0.9615}=2.060\sqrt{1.1096}\approx2.060\times1.0534\approx2.1700$.

Step4: Calculate the confidence interval

The confidence interval for $\mu_1-\mu_2$ is $(\bar{x}_1-\bar{x}_2 - E,\bar{x}_1-\bar{x}_2 + E)$.
Substitute $\bar{x}_1-\bar{x}_2 = 2$ and $E = 2.1700$.
The lower bound is $2-2.1700=-0.1700$.
The upper bound is $2 + 2.1700=4.1700$.

Answer:

$-0.1700<\mu_1-\mu_2<4.1700$