Body Temperatures Refer to Data Set 3 “Body Temperatures” in Appendix B and use allof the matched pairs of body temperatures at 8 AM and 12 AM on Day 1. When using a 0.05significance level for testing a claim of a difference between the temperatures at 8 AM and at12 AM on Day 1, how are the hypothesis test results and confidence interval results affected if the temperatures are converted from degrees Fahrenheit to degrees Celsius? What is therelationship between the confidence interval limits for the body temperatures in degreesFahrenheit and the confidence interval limits for the body temperatures in degrees Celsius?

Hint: \(C = \frac{5}{9}\left( {F - 32} \right)\)

The data of body temperatures (in degrees Fahrenheit) at 8 AM and 12 AM on Day 1is

The level of significance is 0.05.

Null hypothesis: The mean ofthedifferences between the temperatures at 8 AM and at 12 AM on Day 1 is equal to 0.

\({H_0}:{\mu _d} = 0\)

Alternate hypothesis: The mean ofdifference between the temperatures at 8 AM and at 12 AM on Day 1 is not equal to 0.

\({\rm{ }}{H_1}:{\mu _d} \ne 0\)

The following table shows the differences in the temperatures between 8AM and 12AM:

The sample mean of the differences is computed below:

\(\begin{array}{c}\bar x = \frac{{0 + \left( { - 0.6} \right) + ..... + \left( { - 0.2} \right)}}{{17}}\\ = - 0.447\end{array}\)

The sample standard deviation of the differences is computed below:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {0 - \left( { - 0.447} \right)} \right)}^2} + {{\left( {\left( { - 0.6} \right) - \left( { - 0.447} \right)} \right)}^2} + ...... + {{\left( {\left( { - 0.2} \right) - \left( { - 0.447} \right)} \right)}^2}}}{{17 - 1}}} \\ = 0.712\end{array}\)

The value of the test statistic is computed below:

\(\begin{array}{c}t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{\left( { - 0.447} \right) - 0}}{{\frac{{0.712}}{{\sqrt {17} }}}}\\ = - 2.587\end{array}\)

Thus, t is equal to -2.587.

The degrees of freedom is equal to:

\(\begin{array}{c}n - 1 = 17 - 1\\ = 16\end{array}\)

Referring to the t-distribution table, the critical values of t at\(\alpha = 0.05\) with 16 degrees of freedom for a two-tailed test are -2.1199 and 2.1199.

Referring to the t-distribution table, the p-value of t equal to -2.587 is equal to 0.0199.

Since the test statistic value does not lie within the critical values, and the p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to conclude that there is a difference in the mean temperatures at 8AM and 12 AM.

The value of \({t_{\frac{\alpha }{2}}}\) at \(\alpha = 0.05\) is equal to 2.1199 (single-tailed value).

The value of the margin of error is computed as follows:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}\frac{{{s_d}}}{{\sqrt n }}\\ = {t_{\frac{{0.05}}{2}}}\frac{{0.712}}{{\sqrt {17} }}\\ = 2.1199\frac{{0.712}}{{\sqrt {17} }}\\ = 0.3663\end{array}\)

The confidence interval estimate is as follows:

\(\begin{array}{c}CI = \left( {\bar x - E,\bar x + E} \right)\\ = \left( { - 0.447 - 0.3663, - 0.447 + 0.3663} \right)\\ = \left( { - 0.813, - 0.081} \right)\end{array}\)

The 95% confidence interval estimate is equal to (-0.813, -0.081).

Since the interval does not contain 0, there is a significant difference between the mean temperatures at 8 AM and 12 AM.

Use the given expression to convert the temperatures to degrees Celsius:

\(C = \frac{5}{9}\left( {F - 32} \right)\)

Here, F is the temperature value in degrees Fahrenheit.

The value of the temperatures at 8 AM are converted to degrees Celsius as follows:

\(\begin{array}{c}{C_1} = \frac{5}{9}\left( {{F_1} - 32} \right)\\ = \frac{5}{9}\left( {98 - 32} \right)\\ = 36.67\\{C_2} = \frac{5}{9}\left( {{F_1} - 32} \right)\\ = \frac{5}{9}\left( {97 - 32} \right)\\ = 36.11\\.\\.\\.\\.\\{C_{17}} = \frac{5}{9}\left( {{F_{17}} - 32} \right)\\ = \frac{5}{9}\left( {98 - 32} \right)\\ = 36.67\end{array}\)

The value of the temperatures at 8AM are converted to degrees Celsius as follows:

\(\begin{array}{c}{C_1} = \frac{5}{9}\left( {{F_1} - 32} \right)\\ = \frac{5}{9}\left( {98 - 32} \right)\\ = 36.67\\{C_2} = \frac{5}{9}\left( {{F_1} - 32} \right)\\ = \frac{5}{9}\left( {97.6 - 32} \right)\\ = 36.44\\.\\.\\.\\.\\{C_{17}} = \frac{5}{9}\left( {{F_{17}} - 32} \right)\\ = \frac{5}{9}\left( {98.2 - 32} \right)\\ = 36.78\end{array}\)

The following table shows the temperatures in degrees Celsius:

The following table shows the differences in the temperatures between 8AM and 12AM:

The sample mean of the differences is computed below:

\(\begin{array}{c}\bar x = \frac{{0 + \left( { - 0.33} \right) + ..... + \left( { - 0.11} \right)}}{{17}}\\ = - 0.25\end{array}\)

The sample standard deviation of the differences is computed below:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {0 - \left( { - 0.25} \right)} \right)}^2} + {{\left( {\left( { - 0.33} \right) - \left( { - 0.25} \right)} \right)}^2} + ...... + {{\left( {\left( { - 0.11} \right) - \left( { - 0.25} \right)} \right)}^2}}}{{17 - 1}}} \\ = 0.396\end{array}\)

The value of the test statistic is computed below:

\(\begin{array}{c}t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{\left( { - 0.25} \right) - 0}}{{\frac{{0.396}}{{\sqrt {17} }}}}\\ = - 2.587\end{array}\)

Thus, t is equal to -2.587.

The degrees of freedom is equal to:

\(\begin{array}{c}n - 1 = 17 - 1\\ = 16\end{array}\)

Referring to the t-distribution table, the critical values of t at\(\alpha = 0.05\) with 16 degrees of freedom for a two-tailed test are -2.1199 and 2.1199.

Referring to the t-distribution table, the p-value of t equal to -2.587 is equal to 0.0199.

Since the test statistic value does not lie within the critical values and the p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to conclude that there is a difference in the mean temperatures at 8AM and 12 AM.

The value of \({t_{\frac{\alpha }{2}}}\) at \(\alpha = 0.05\) is equal to 2.1199 (single-tailed value).

The value of the margin of error is computed as follows:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}\frac{{{s_d}}}{{\sqrt n }}\\ = {t_{\frac{{0.05}}{2}}}\frac{{0.396}}{{\sqrt {17} }}\\ = 2.1199\frac{{0.396}}{{\sqrt {17} }}\\ = 0.2035\end{array}\)

The confidence interval estimate is as follows:

\(\begin{array}{c}CI = \left( {\bar x - E,\bar x + E} \right)\\ = \left( { - 0.25 - 0.2035, - 0.25 + 0.2035} \right)\\ = \left( { - 0.452, - 0.045} \right)\end{array}\)

The 95% confidence interval estimate is equal to (-0.452, -0.045).

Since the interval does not contain 0, there is a significant difference between the mean temperatures at 8 AM and 12 AM.

The results of the hypothesis test and the confidence interval, are the same for the temperatures in degrees Fahrenheit and degrees Celsius.

Lower Limit:

\(\frac{5}{9}\left( { - 0.813} \right) = - 0.452\)

Upper Limit:

\(\frac{5}{9}\left( { - 0.081} \right) = - 0.045\)

As shown above, the confidence interval limits corresponding to the temperatures in \(\left( {^ \circ C} \right)\) are 5/9 times the confidence interval limits corresponding to the temperatures in \(\left( {^ \circ F} \right)\).