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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 \(\left( {^ \circ F} \right)\).