Question

A report included the following information on the heights (in.) for non-Hispanic white females.

Answer 1

Answer:

Following are the solution to the given points:

Step-by-step explanation:

In point a:

When 95% are confidence interval:

$=( \mu_{20-39} - \mu_{60})+(-z_{0.025})((0.09)^2+(0.11)^2)^{0.5} \\\\= (65.9-64.3)+(-1.96) \times 0.14213\\\\=(1.3214, 1.8786)$

In point b:

$H_0: \mu_1 - \mu_2 =1\\\\ H_a: \mu_1 - \mu_2 > 1$

In the Z test statistic:

$= \frac{((65.9-64.3)-1)}{((0.09)^2+(0.11)^2)^{0.5}} \\\\= 4.22 \\\\\to p= 0.000012$

In point c:

The rejection value of the null hypothesis as $p< 0.001$

In point d:

$H_0: \mu_2 -\mu_1 = 1 \\\\H_a: \mu_2 - \mu_1 > 1$