A report included the following information on the heights (in.) for non-Hispanic white females.
Sample Sample Std. Error
Age Size Mean Mean
20–39 868 65.9 0.09
60 and older 934 64.3 0.11
(a) Calculate a confidence interval at confidence level approximately 95% for the difference between population mean height for the younger women and that for the older women. (Use μ20–39 − μ60 and older.) , Interpret the interval
(b) Let μ1 denote the population mean height for those aged 20–39 and μ2 denote the population mean height for those aged 60 and older. Interpret the hypotheses H0: μ1 − μ2 = 1 and Ha: μ1 − μ2 > 1.
Carry out a test of these hypotheses at significance level 0.001. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)
(c) Based on the P-value calculated in (b) would you reject the null hypothesis at any reasonable significance level? Explain your reasoning.
(d) What hypotheses would be appropriate if μ1 referred to the older age group, μ2 to the younger age group, and you wanted to see if there was compelling evidence for concluding that the population mean height for younger women exceeded that for older women by more than 1 in.?

Respuesta :

Following are the solution to the given points:

In point a:

When 95% are confidence interval:

[tex]=( \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)[/tex]

In point b:

[tex]H_0: \mu_1 - \mu_2 =1\\\\ H_a: \mu_1 - \mu_2 > 1[/tex]

In the Z test statistic:

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

In point c:

The rejection value of the null hypothesis as [tex]p< 0.001[/tex]

In point d:

[tex]H_0: \mu_2 -\mu_1 = 1 \\\\H_a: \mu_2 - \mu_1 > 1[/tex]

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