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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.

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

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. \(\hat p = \frac{x}{n}\)

The likelihood of exactly\(x\)successes on\(n\)repeated trials in an experiment with two alternative outcomes is known as binomial probability (commonly called a binomial experiment).

The binomial probability is\({}_n{C_x} \cdot {p^x} \cdot {(1 - p)^{n - x}}\)if the likelihood of success on an individual trial is\(p\).

The number of alternative combinations of \(x\) objects chosen from a set of \(n\) objects is indicated by \({}_n{C_x}\).

Let the given be: The probability of rejecting the null hypothesis is at most\(0.10\).

\(\alpha = 0.10\)

The probability of failing to reject the null hypothesis is at most\(0.10\).

\(\beta (0.30) = 0.10\)

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.

\(\begin{array}{l}{H_0}:p = 0.10\\{H_a}:p > 0.10\end{array}\)

Using the normal probability table in the appendix, find the z-score corresponding to a probability of\(1 - \alpha = 0.90\)(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:

\(\begin{aligned}{c}\hat p &= p + z\sqrt {\frac{{p(1 - p)}}{n}} \\ &= 0.10 + 1.28\sqrt {\frac{{0.10(1 - 0.10)}}{n}} \end{aligned}\)

Using the normal probability table in the appendix, find the z-score corresponding to a probability of\(\beta (0.30) = 0.10\)(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:

\(\begin{aligned}{c}\hat p &= p + z\sqrt {\frac{{p(1 - p)}}{n}} \\ &= 0.30 - 1.28\sqrt {\frac{{0.30(1 - 0.30)}}{n}} \end{aligned}\)

The both expressions must be equal to each other.

\(\begin{aligned}{c}0.10 + 1.28\sqrt {\frac{{0.10(1 - 0.10)}}{n}} &= 0.30 - 1.28\sqrt {\frac{{0.30(1 - 0.30)}}{n}} \\ - 0.20 + 1.28\sqrt {\frac{{0.10(1 - 0.10)}}{n}} &= - 1.28\sqrt {\frac{{0.30(1 - 0.30)}}{n}} \\ - 0.20 &= - 1.28\sqrt {\frac{{0.30(1 - 0.30)}}{n}} - 1.28\sqrt {\frac{{0.10(1 - 0.10)}}{n}} \end{aligned}\)

\(\begin{aligned}{c} - 0.20\sqrt n &= - 1.28\sqrt {0.30(1 - 0.30)} - 1.28\sqrt {0.10(1 - 0.10)} \\\sqrt n &= \frac{{1.28\sqrt {0.30(1 - 0.30)} + 1.28\sqrt {0.10(1 - 0.10)} }}{{0.20}}\\n &= {\left( {\frac{{1.28\sqrt {0.30(1 - 0.30)} + 1.28\sqrt {0.10(1 - 0.10)} }}{{0.20}}} \right)^2}\\ &\approx 24\end{aligned}\)

Let solve for \(\alpha \):

The highest possible probability is \(\alpha = P(X > 4) = 0.0980\).

Let solve for\(\beta \):

The highest possible probability is \(\begin{array}{c}\beta = P(X \le 4) = P(X = 0) + P(X = 1) + ... + P(X = 4)\\ = 0.0905\end{array}\).