91Ó°ÊÓ

a)

For sufficiently big m and n, the confidence interval for \({\mu _1} - {\mu _2}\) with a confidence level of about \(100\left( {1 - \alpha } \right)\% \) percent is.

\(\bar x - \bar y \pm {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \)

An upper/lower bound is obtained by replacing \({z_{\alpha /2}}\) with \( {z_\alpha }\) and \( \pm \) with only \( + \) and \( - \).

The first sample has a sample size of \(m = 866 \), while the second sample has a sample size of \(n = 934\), both of which are large enough to employ the specified confidence interval.

For the first sample \((20 - 39)\), the standard error of the mean is,

\(\frac{{{s_1}}}{{\sqrt m }} = 0.09\)

while for the second sample (age 60 and older), it is.

\(\frac{{{s_2}}}{{\sqrt n }} = 0.11\)

It is necessary to obtain alpha and then \({z_{\alpha /2}}\) in order to obtain a confidence level of around \(95\% \).

\(100 \times (1 - \alpha ) = 95\)

\(\alpha = 0.05\) is the result of equation.

\({z_{\alpha /2}} = {z_{0.025}} = 1.96\)

As a result, the \(z\) value comes from the table in the book's appendix.

\(\bar x = 64.9,\;\;\;\bar y = 63.1\)

Finally, for \({\mu _1} - {\mu _2}\), the \(95\% \) percent confidence interval can be calculated. This is the lower limit.

\(\begin{array}{l}\bar x - \bar y - {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} = 64.9 - 63.1 - 1.96 \times \sqrt {0.0{9^2} + 0.1{1^2}} \\ = 1.8 - 0.2786\\ = 1.5214.\end{array}\)

This is the upper limit.

\(\begin{array}{l}\bar x - \bar y + {z_{\alpha /2}} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} = 64.9 - 63.1 + 1.96 \times \sqrt {0.0{9^2} + 0.1{1^2}} \\ = 1.8 + 0.2786\\ = 2.0786\end{array}\)

For \({\mu _1} - {\mu _2}\), the \(95\% \) percent confidence interval is.

\((1.5214,2.0786)\)

b)

The hypothesis \({H_0}:{\mu _1} - {\mu _2} = 1\) can be translated as follows: the difference between the sample means is equal to one. The hypothesis \({H_1}:{\mu _1} - {\mu _2} > 1\) can be translated as: the difference in sample means is greater than \(1\).

When there is a null hypothesis.

\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)

The test statistic value

\(z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}.\)

The $P$ value is derived for the acceptable alternative hypothesis by calculating the adequate area under the standard normal curve. When both \(m > 40 and n > 40\), this test can be employed.

Because both \(m\) and \(n\) are greater than \(40\), the test statistic described above can be utilized. The test statistic's value is.

\(\begin{array}{l}z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{64.9 - 63.1 - 1}}{{\sqrt {0.0{9^2} + 0.1{1^2}} }} = \frac{{0.8}}{{0.1421}}\\ = 5.63.\end{array}\)

The \(P\) value is the area under the standard normal curve to the right of \(Z\) in this one-sided test. The value of \(P\) is.

\(P(Z > 5.63) \approx 0\)

At any reasonable level

The null hypothesis is reject

Since

\(P < \alpha = 0.001\)

There is evidence to suggest that the difference in means is more than one.

c)At any appropriate level, as stated in (b)

Reject the null hypothesis.

Since

\(P < \alpha \)

For any reasonable \(\alpha \)

d) \({H_0}\)would be the null hypothesis.

\({H_0}:{\mu _1} - {\mu _2} = - 1\left( {{\mu _2} > {\mu _1} by 1 in. } \right) \)versus \({H_a}:{\mu _1} - {\mu _2} < - 1\left( {{\mu _2}} \right.\)exceeds \({\mu _1}\) by more than \( 1 in.\)).

a) \((1.5214,2.0786)\)

b) reject null hypothesis;

c) reject the null hypothesis at any reasonable significance level;

d) \({H_0}:{\mu _1} - {\mu _2} = - 1 versus {H_a}:{\mu _1} - {\mu _2} < - 1\)