91Ó°ÊÓ

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.

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

The formula for the value of the test statistic is given by,\(z = \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\).

The sample proportion is calculated by dividing the number of successes by the sample size. \(\hat p = \frac{x}{n}\)

(a)

Let the given be:

\(\begin{array}{c}x = 14\\n = 100\\\alpha = 0.05\end{array}\)

Claim that the proportion is more than\(0.10\)or\(10\% \).

Sample proportion:

\(\begin{aligned}{c}\hat p &= \frac{x}{n}\\ &= \frac{{14}}{{100}}\\ &\approx 0.14\end{aligned}\)

The value of the test-statistic:

\(\begin{aligned}{c}z &= \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\\ &= \frac{{0.14 - 0.10}}{{\sqrt {\frac{{0.10(1 - 0.10)}}{{100}}} }}\\ &\approx 1.33\end{aligned}\)

When the null hypothesis is true, the P-value is the chance of getting the test statistic's value, or a value that is more extreme. Using the normal probability table in the appendix, calculate the P-value.

\(\begin{aligned}{c}P &= P(Z > 1.33)\\ &= 1 - P(Z < 1.33)\\ &= 1 - 0.9082\\ &= 0.0918\end{aligned}\)

Since the P-value is smaller than the significance level\(\alpha \), then reject the null hypothesis:

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

There isn't enough evidence to back up the allegation that more than \(10\% \) of all plates blister under such circumstances.

(b)

Let: \(\begin{array}{l}{p_A} = 15\% = 0.15\\n = 100\end{array}\)

Claim that the proportion is more than\(0.10\)or\(10\% \).

Using the normal probability table in the appendix, find the z-score corresponding to a probability of\(1 - \alpha = 0.95\)(Note: take the complement because the test is right-sided):

\(z = 1.645\)

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

\(\begin{array}{c}\hat p = p + z\sqrt {\frac{{p(1 - p)}}{n}} \\ = 0.10 + 1.645\sqrt {\frac{{0.10(1 - 0.10)}}{{100}}} \\ \approx 0.14935\end{array}\)

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

\(\begin{array}{c}z = \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\\ = \frac{{0.14935 - 0.15}}{{\sqrt {\frac{{0.15(1 - 0.15)}}{{100}}} }}\\ \approx - 0.02\end{array}\)

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.

\(\begin{aligned}{c}\beta &= P(Z < - 0.02)\\ &= 0.4920\\ &= 49.20\% \end{aligned}\)

Now,\(n = 200\).

Using the normal probability table in the appendix, find the z-score corresponding to a probability of\(1 - \alpha = 0.95\)(Note: take the complement because the test is right-sided):

\(z = 1.645\)

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

\(\begin{aligned}{c}\hat p &= p + z\sqrt {\frac{{p(1 - p)}}{n}} \\ &= 0.10 + 1.645\sqrt {\frac{{0.10(1 - 0.10)}}{{200}}} \\ &\approx 0.1349\end{aligned}\)

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

\(\begin{aligned}{c}z &= \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\\ &= \frac{{0.1349 - 0.15}}{{\sqrt {\frac{{0.15(1 - 0.15)}}{{200}}} }}\\ &\approx - 0.60\end{aligned}\)

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.

\(\begin{aligned}{c}\beta &= P(Z < - 0.60)\\ &= 0.2743\\ &= 27.43\% \end{aligned}\)

(c)

Now,\(\beta (0.15) = 0.10\).

Using the normal probability table in the appendix, find the z-score corresponding to a probability of\(1 - \alpha = 0.95\)(Note: take the complement because the test is right-sided):

\(z = 1.645\)

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

\(\begin{aligned}{c}\hat p &= p + z\sqrt {\frac{{p(1 - p)}}{n}} \\ &= 0.10 + 1.645\sqrt {\frac{{0.10(1 - 0.10)}}{n}} \end{aligned}\)

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:

\(\begin{aligned}{c}\hat p &= p + z\sqrt {\frac{{p(1 - p)}}{n}} \\ &= 0.15 - 1.28\sqrt {\frac{{0.15(1 - 0.15)}}{n}} \end{aligned}\)

The two-sample expression must be equal to each other. Solve for \(n\):

\(\begin{aligned}{c}0.10\sqrt n + 1.645\sqrt {0.10(1 - 0.10)} &= 0.15\sqrt n - 1.28\sqrt {0.15(1 - 0.15)} \\ - 0.05\sqrt n + 1.645\sqrt {0.10(1 - 0.10)} &= - 1.28\sqrt {0.15(1 - 0.15)} \\ - 0.05\sqrt n &= - 1.28\sqrt {0.15(1 - 0.15)} - 1.645\sqrt {0.10(1 - 0.10)} \\\sqrt n &= \frac{{ - 1.28\sqrt {0.15(1 - 0.15)} - 1.645\sqrt {0.10(1 - 0.10)} }}{{ - 0.05}}\\n &= {\left( {\frac{{ - 1.28\sqrt {0.15(1 - 0.15)} - 1.645\sqrt {0.10(1 - 0.10)} }}{{ - 0.05}}} \right)^2}\\ &\approx 362\end{aligned}\)