Question

Given two independent random samples with the following results: n1=8x‾1=186s1=33 n2=7x‾2=171s2=23 Use this data to find the 90% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval.

Answer 1

Answer:

The point of estimate for the true difference would be:

$186-171= 15$

And the confidence interval is given by:

$(186-171) -1.77 \sqrt{\frac{33^2}{8} +\frac{23^2}{7}}= -10.753$

$(186-171) +1.77 \sqrt{\frac{33^2}{8} +\frac{23^2}{7}}= 40.753$

Step-by-step explanation:

For this case we have the following info given:

$\bar X_1 = 186$ the sample mean for the first sample

$\bar X_2 = 171$ the sample mean for the second sample

$s_1 =33$ the sample deviation for the first sample

$s_2 =23$ the sample deviation for the second sample

$n_1 = 8$ the sample size for the first group

$n_2 = 7$ the sample size for the second group

The confidence interval for the true difference is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

We can find the degrees of freedom are given:

$df = n_1 +n_2 -2 =8+7-2= 13$

The confidence level is given by 90% so then the significance would be $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ we can find the critical value with the degrees of freedom given and we got:

$t_{\alpha/2}= \pm 1.77$

The point of estimate for the true difference would be:

$186-171= 15$

And replacing into the formula for the confidence interval we got:

$(186-171) -1.77 \sqrt{\frac{33^2}{8} +\frac{23^2}{7}}= -10.753$

$(186-171) +1.77 \sqrt{\frac{33^2}{8} +\frac{23^2}{7}}= 40.753$