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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.