91Ó°ÊÓ

Given:

\(\begin{array}{l}\bar d = 3.75\\{s_d}/\sqrt n = 3.878\\n = 41\\c = 95\% = 0.95\end{array}\)

Determine the\({t_{\alpha /2}}\) using the Student's T distribution table in the appendix with \(df = n - 1 = 41 - 1 = 40\) :

\({t_{0.025}} = 2.021\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }} = 2.021 \cdot 3.878 = 7.837438\)

The endpoints of the confidence interval for d are:

\(\begin{array}{l}\bar d - E = 3.75 - 7.837438 = - 4.087438\\\bar d + E = 3.75 + 7.837438 = 11.587438\end{array}\)

Given:

\(\begin{array}{l}\bar d = 9.05\\{s_d}/\sqrt n = 4.256\\n = 36\end{array}\)

Let us assume:

\(\alpha = 0.05\)

Given claim: increases

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.

\(\begin{array}{l}{H_0}:{\mu _d} = 0\\{H_a}:{\mu _d} > 0\end{array}\)

Determine the value of the test statistic:

\(t = \frac{{\bar d - d}}{{{s_d}/\sqrt n }} = \frac{{9.05 - 0}}{{4.256}} \approx 2.126\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The Pvalue is the number (or interval) in the column title of the Student's T distribution in the appendix containing the t-value in the row\(d{\bf{f}} = {\bf{n}} - {\bf{1}} = \)\(36 - 1 = 35 > 34\) :

\(0.01 < P < 0.025\)

If the P-value is less than the significance level, reject the null hypothesis.

\(P < 0.05 \Rightarrow {\mathop{\rm Reject}\nolimits} {H_0}\)

There is sufficient evidence to support the claim that the true average cholesterol level increases at the \(0.05\) significance level.

The choice of significance level impacts your conclusion, because you will make a different conclusion at the \(0.01\) significance level.

\(\begin{array}{l}c = 99\% = 99\\n = 45\end{array}\)

\(95\% \)confidence interval: \((7.38,9.69)\)

The sample mean is the center of the confidence interval and thus is the average of the boundaries of the confidence interval:

\(\bar d = \frac{{7.38 + 9.69}}{2} = 8.535\)

Determine the \({t_{\alpha /2}}\) using the Student's T distribution table in the appendix with\(df = n - 1 = 45 - 1 = 44 > 40\):

\({t_{0.025}} = 2.021\)

The margin of error is given by the formula:

\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }}\)

The margin of error is half the width of the confidence interval:

\(\frac{{9.69 - 7.38}}{2} = 2.021 \cdot \frac{{{s_d}}}{{\sqrt {45} }}\)

Divide each side by \(2.021/\sqrt {45} \) :

\({s_d} = \frac{{(9.69 - 7.38)\sqrt {45} }}{{2(2.021)}} \approx 3.8337\)

Determine the\({t_{\alpha /2}}\)using the Student's T distribution table in the appendix with\(df = n - 1 = 45 - 1 = 44 > 40\):

\({t_{0.005}} = 2.704\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }} = 2.704 \cdot \frac{{3.8337}}{{\sqrt {45} }} = 1.5453\)

The endpoints of the confidence interval for\({\mu _d}\)are:

\(\begin{array}{l}\bar d - E = 8.535 - 1.5453 = 6.9897\\\bar d + E = 8.535 + 1.5453 = 10.0803\end{array}\)