91影视

Let us assume:

\(\alpha = 0.05\)

Determine the difference in value of each pair.

Given claim: a change

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.

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

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{ - 1 + 2 + 24 + 35 - 16 - 9}}{6} \approx 5.8333\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{( - 1 - 5.8333)}^2} + \ldots + {{( - 9 - 5.8333)}^2}}}{{6 - 1}}} \approx 19.6918\)

Determine the value of the test statistic:

\(t = \frac{{\bar d - d}}{{{s_d}/\sqrt n }} = \frac{{5.8333 - 0}}{{19.6918/\sqrt 6 }} \approx 0.726\)

The\({\rm{P}}\)-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 \(df = n - 1 = 6 - 1 = 5\) :

\(P > 2 \times 0.10 = 0.20\)If the P-value is less than the significance level, reject the null hypothesis.

\(P > 0.05 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

There is not sufficient evidence to support the claim that there was any change in the mean number of crashes before and after the addition of distance information on the signs.

We will determine a\(95\% \) prediction interval. Other confidence levels can be determined similarly.

\(c = 95\% = 0.95\)

Determine the difference in value of each pair.

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{ - 1 + 2 + 24 + 35 - 16 - 9}}{6} \approx 5.8333\)

Determine the sample standard deviation of the differences: \({s_d} = \sqrt {\frac{{{{( - 1 - 5.8333)}^2} + \ldots + {{( - 9 - 5.8333)}^2}}}{{6 - 1}}} \approx 19.6918\)

Determine the t-value by looking in the row starting with degrees of freedom\(df = n - 1 = 6 - 1 = 5\)and in the column with\(\alpha = 1 - c/2 = 0.025\)in the Student's t distribution table in the appendix:

\({t_{\alpha /2}} = 2.571\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot {s_d}\sqrt {1 + \frac{1}{n}} = 2.571 \cdot 19.6918\sqrt {1 + \frac{1}{6}} \approx 54.6841\)

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

\(\begin{array}{l}\bar d - E = 5.8333 - 54.6841 = - 48.8508\\\bar d + E = 5.8333 + 54.6841 = 60.5174\end{array}\)