Body TemperaturesListed below are body temperatures from seven different subjects measuredat two different times in a day (from Data Set 3 “Body Temperatures” in Appendix B).

a.Use a 0.05 significance level to test the claim that there is no difference between body temperaturesmeasured at 8 AM and at 12 AM.

b.Construct the confidence interval that could be used for the hypothesis test described in part(a). What feature of the confidence interval leads to the same conclusion reached in part (a)

Body Temperature\(\left( {^{\bf{0}}{\bf{F}}} \right)\) at 8AM	96.6	97.0	97.0	97.8	97.0	97.4	96.6
Body Temperature\(\left( {^{\bf{0}}{\bf{F}}} \right)\) at 12AM	99.0	98.4	98.0	98.6	98.5	98.9	98.4

The body temperatures of a set of 7 subjects are recorded, first at 8 AM and then at 12 AM. Therefore, the sample size is equal to \(n = 5\). It is claimed that there is no difference between the body temperatures measured at 8 AM and 12 AM.

The following hypotheses are noted:

Null Hypothesis: The difference between the body temperatures measured at 8 AM and 12 AM is equal to 0.

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

Alternative Hypothesis: The difference between the body temperatures measured at 8 AM and 12 AM is not equal to 0.

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

Since the alternative hypothesis contains a not equal sign, it is a two-tailed test.

The following table shows the differences in temperatures for each pair:

Themean value of the differences between thebody temperatures at 8 AM and 12 AM:

\(\begin{array}{c}\bar d = \frac{{\sum d }}{n}\\ = \frac{{ - 10.4}}{7}\\ = - 1.49\end{array}\)

Thus, the mean value is -1.49 degrees Fahrenheit.

Thestandard deviation of thedifferences between the body temperaturesis:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{d_i} - \bar d} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( { - 2.4 - \left( { - 1.49} \right)} \right)}^2} + {{\left( { - 1.4 - \left( { - 1.49} \right)} \right)}^2} + ...... + {{\left( { - 1.8 - \left( { - 1.49} \right)} \right)}^2}}}{{7 - 1}}} \\ = 0.52\end{array}\)

The standard deviation is 0.52 degrees Fahrenheit.

The test statistic for matched pair is equal to:

\(\begin{array}{c}t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{ - 1.49 - 0}}{{\frac{{0.52}}{{\sqrt 7 }}}}\\ = - 7.50\end{array}\)

Thus, t=-7.50.

The value of the degrees of freedom is equal to:

\(\begin{array}{c}df = n - 1\\ = 7 - 1\\ = 6\end{array}\)

Referring to the t-distribution table, use the column for 0.05 (Area in One Tail),and use the row with 6 degrees of freedom to find the critical value.

\(\begin{array}{c}{t_{crit}} = {t_\alpha },df\\ = {t_{0.05}},6\\ = \left( { - 2.4469,2.4469} \right)\end{array}\)

Thus, the critical value is (-2.4469, 2.4469).

a.

For a two-tailed test, reject the null hypothesis if the calculated test statistic does not lie in the interval of critical values.

Here -7.58 does not lie in the interval -2.45 and +2.45.

Hence reject the null hypothesis.

There is sufficient evidence to reject the claim that the differences between the body temperatures measured between 8 AM and 12 AM have a mean value equal to 0.

If the level of significance used in the one-tailed hypothesis test is equal to 0.04, then construct a 90% confidence interval.

The margin of error is equal to:

\(\)\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \frac{{{s_d}}}{{\sqrt n }}\\ = 2.4469 \times \frac{{0.52}}{{\sqrt 7 }}\\ = 0.485\end{array}\)

b.

The formula for the confidence interval is equal to:

\(\begin{array}{c}CI = \bar d - E < {\mu _d} < \bar d + E\\ = - 1.49 - 0.485 < {\mu _d} < - 1.49 + 0.485\\ = - 1.970 < {\mu _d} < - 1.001\end{array}\)

Thus, the confidence interval is equal to (-1.970, -1.001).

Here clearly, one can see that 0 is not included in the interval.

Therefore, the mean differences cannot be equal to zero.

Hencethere is sufficient evidence to reject the claim that there is a difference between the body temperatures measured between 8 AM and 12 AM have a mean equal to 0.