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The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

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

The formula for the value of the test statistic is given by, \(t = \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\).

The mean is the ration of sum of all values and the total number of values.

\(\begin{array}{c}\bar x = \frac{{32.1 + 30.6 + 31.4 + 30.4 + 31 + 31.9}}{6}\\ = \frac{{187.4}}{6}\\ \approx 31.2333\end{array}\)

The square of the variance is the standard deviation.

\(\begin{array}{c}s = \sqrt {\frac{{{{(32.1 - 31.2333)}^2} + \ldots . + {{(31.9 - 31.2333)}^2}}}{{6 - 1}}} \\ \approx 0.6890\end{array}\)

(a)

Let the given be:

\(\begin{array}{l}n = 6\\\alpha = 0.01\end{array}\)

Claim that the average stopping distance exceeds\(30\). The given claim is either the null hypothesis or the alternative hypothesis.

The value of the test statistic:

\(\begin{array}{c}t = \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\\ = \frac{{31.2333 - 30}}{{0.6890/\sqrt 6 }}\\ \approx 4.385\end{array}\)

The P-value is the chance of getting the test statistic's result, or a number that is more severe. The P-value is the number (or interval) in the column header of the T table in the appendix that contains the t-value in the row\(\begin{array}{c}df = n - 1\\ = 6 - 1\\ = 5\end{array}\)for the student.

\(0.001 < P < 0.005\)

As the P-value is smaller than the significance level, so the null hypothesis is rejected.

\(P < 0.01 \Rightarrow {\rm{Reject }}{H_0}\)

Hence, the average stopping distance exceeds \(30\) is supported by appropriate evidence.

(b)

Let the given be:

\(\begin{array}{l}\sigma = 0.65\\n = 6\\\alpha = 0.01\end{array}\)

Claim that the average stopping distance exceeds\(30\). The given claim is either the null hypothesis or the alternative hypothesis.

\({\mu _A} = 31\)

Using the normal probability table in the appendix, find the z-score corresponding to a probability of \(1 - \alpha = 0.99\) (Note: take the complement because the test is right-sided):

\(z = 2.33\)

The population mean (of the hypothesis) is increased by the product of the z-score and the standard deviation to get the sample mean:

\(\begin{array}{c}\bar x = \mu + z\frac{\sigma }{{\sqrt n }}\\ = 30 + 2.33\frac{{0.65}}{{\sqrt 6 }}\\ \approx 30.6183\end{array}\)

The z-value is the sample mean divided by the standard deviation, after subtracting the population mean (alternative mean).

\(\begin{array}{c}z = \frac{{\bar x - \mu }}{{\sigma /\sqrt n }}\\ = \frac{{30.6183 - 31}}{{0.65/\sqrt 6 }}\\ \approx - 1.44\end{array}\)

When the null hypothesis is false, the probability of making a type II error is the probability of not rejecting the null hypothesis. Using the normal probability table in the appendix, calculate the chances of failing to reject the null hypothesis.

\(\begin{array}{c}P({\rm{ Type II error }}) = P(Z < - 1.44)\\ = 0.0749\\ = 7.49\% \end{array}\)

Now,\({\mu _A} = 32.\)

The z-value is the sample mean divided by the standard deviation, after subtracting the population mean (alternative mean).

\(\begin{array}{c}z = \frac{{\bar x - \mu }}{{\sigma /\sqrt n }}\\ = \frac{{30.6183 - 32}}{{0.65/\sqrt 6 }}\\ \approx - 5.21\end{array}\)

When the null hypothesis is false, the probability of making a type II error is the probability of not rejecting the null hypothesis. Using the normal probability table in the appendix, calculate the chances of failing to reject the null hypothesis.

\(\begin{array}{c}P({\rm{ Type II error }}) = P(Z < - 5.21)\\ \approx 0\end{array}\)

(c)

Let the given be:

\(\begin{array}{l}\sigma = 0.80\\n = 6\\\alpha = 0.01\end{array}\)

Claim that the average stopping distance exceeds\(30\). The given claim is either the null hypothesis or the alternative hypothesis.

\({\mu _A} = 31\)

Using the normal probability table in the appendix, find the z-score corresponding to a probability of \(1 - \alpha = 0.99\) (Note: take the complement because the test is right-sided):

\(z = 2.33\)

The population mean (of the hypothesis) is increased by the product of the z-score and the standard deviation to get the sample mean:

\(\begin{array}{c}\bar x = \mu + z\frac{\sigma }{{\sqrt n }}\\ = 30 + 2.33\frac{{0.80}}{{\sqrt 6 }}\\ \approx 30.7610\end{array}\)

The z-value is the sample mean divided by the standard deviation, after subtracting the population mean (alternative mean).

\(\begin{array}{c}z = \frac{{\bar x - \mu }}{{\sigma /\sqrt n }}\\ = \frac{{30.7610 - 31}}{{0.80/\sqrt 6 }}\\ \approx - 0.73\end{array}\)

When the null hypothesis is false, the probability of making a type II error is the probability of not rejecting the null hypothesis. Using the normal probability table in the appendix, calculate the chances of failing to reject the null hypothesis.

\(\begin{array}{c}P({\rm{ Type II error }}) = P(Z < - 0.73)\\ = 0.2327\\ = 23.27\% \end{array}\)

Now,\({\mu _A} = 32.\)

The z-value is the sample mean divided by the standard deviation, after subtracting the population mean (alternative mean).

\(\begin{array}{c}z = \frac{{\bar x - \mu }}{{\sigma /\sqrt n }}\\ = \frac{{30.7610 - 32}}{{0.80/\sqrt 6 }}\\ \approx - 3.79\end{array}\)

When the null hypothesis is false, the probability of making a type II error is the probability of not rejecting the null hypothesis. Using the normal probability table in the appendix, calculate the chances of failing to reject the null hypothesis.

\(\begin{array}{c}P({\rm{ Type II error }}) = P(Z < - 3.79)\\ \approx 0\end{array}\)

(d)

Let the given be:

\(\begin{array}{l}{\mu _A} = 31\\\sigma = 0.65\\n = ?\\\alpha = 0.01\\\beta = 0.10\end{array}\)

Claim that the average stopping distance exceeds\(30\). The given claim is either the null hypothesis or the alternative hypothesis.

Using the normal probability table in the appendix, find the z-score corresponding to a probability of \(1 - \alpha = 0.99\) (Note: take the complement because the test is right-sided):

\(z = 2.33\)

The population mean (of the hypothesis) is increased by the product of the z-score and the standard deviation to get the sample mean:

\(\begin{array}{c}\bar x = \mu + z\frac{\sigma }{{\sqrt n }}\\ = 30 + 2.33\frac{{0.65}}{{\sqrt n }}\end{array}\)

Using the normal probability table in the appendix, find the z-score corresponding to a probability of \(\beta = 0.10\) (Note: take the complement because the test is right-sided):

\(z = - 1.28\)

The population mean (of the hypothesis) is increased by the product of the z-score and the standard deviation to get the sample mean:

\(\begin{array}{c}\bar x = \mu + z\frac{\sigma }{{\sqrt n }}\\ = 31 - 1.28\frac{{0.65}}{{\sqrt n }}\end{array}\)

The two expressions for the sample mean must be equal. Solve for\(n\):

\(\begin{array}{c}30 + 2.33\frac{{0.65}}{{\sqrt n }} = 31 - 1.28\frac{{0.65}}{{\sqrt n }}\\30 = 31 - 3.61\frac{{0.65}}{{\sqrt n }}\\ - 1 = - 3.61\frac{{0.65}}{{\sqrt n }}\\ - \sqrt n = - 3.61(0.65)\\\sqrt n = 2.3465\\n = {2.3465^2}\\ = 5.50606225\\n \approx 6\end{array}\)