Question

The National Health Statistics Reports dated Oct. \(22,2008\), included the following information on the heights (in.) for non-Hispanic white females:

Sample sample Std. Error

Age Size Mean Mean

\(\begin{array}{*{20}{l}}{20 - 39}&{866}&{64.9}&{.09}\\{60 and older }&{934}&{63.1}&{.11}\\{}&{}&{}&{}\end{array}\)

1. Calculate and interpret 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.
2. Let \({\mu _1}\) denote the population mean height for those aged \(20 - 39\) and \({\mu _2}\) denote the population mean height for those aged 60 and older. Interpret the hypotheses \({H_0}:{\mu _1} - {\mu _2} = 1 and {H_a}:{\mu _1} - {\mu _2} > 1,\) and then carry out a test of these hypotheses at significance level \(.001\)
3. Based on the \(p\)-value calculated in (b) would you reject the null hypothesis at any reasonable significance level? Explain your reasoning.
4. What hypotheses would be appropriate if \({\mu _1}\) referred to the older age group, \({\mu _2}\) 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.?

Short answer

the solution is

a) \((1.5214,2.0786)\)

b) reject null hypothesis;

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

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

Step-by-step solution

find the population

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.

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

calculate the limit

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.

\((1.5214,2.0786)\)

test the static value

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.

calculate p-value

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 \( 1 in.\)).

conclusion

a) \((1.5214,2.0786)\)

b) reject null hypothesis;

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

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