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$Consider the following hypothesis test. {:[H_(0):mu >= 10],[H_(a):mu < 10]:} The sample size is 105 and the population standard deviation is assumed known with sigma=5. Use alpha=0.05. (a) If the population mean is 9 , what is the probability that the sample mean leads to the conclusion do not reject H_(0) ? (Round your answer to four decimal places.) ◻ (b) What type of error would be made if the actual population mean is 9 and we conclude that H_(0):mu >= 10 is true? Type I Type II (c) What is the probability of making a type II error if the actual population mean is 8? (Round your answer to four decimal places. If it is not possible to commit a type II error enter NOT POSSIBLE.) ◻$

Consider the following hypothesis test. $\newline$ $\begin{array}{l} H_{0}: \mu \geq 10 \\ H_{\mathrm{a}}: \mu<10 \end{array}$ $\newline$ The sample size is $105$ and the population standard deviation is assumed known with $\sigma=5$ . Use $\alpha=0.05$ . $\newline$ (a) If the population mean is $9$ , what is the probability that the sample mean leads to the conclusion do not reject $H_{0}$ ? (Round your answer to four decimal places.) $\newline$ $\square$ $\newline$ (b) What type of error would be made if the actual population mean is $9$ and we conclude that $H_{0}: \mu \geq 10$ is true? $\newline$ Type I $\newline$ Type II $\newline$ (c) What is the probability of making a type II error if the actual population mean is $8$ ? (Round your answer to four decimal places. If it is not possible to commit a type II error enter NOT POSSIBLE.) $\newline$ $\square$

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Algebra 2

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Q. Consider the following hypothesis test.

\newline

\begin{array}{l} H_{0}: \mu \geq 10 \\ H_{\mathrm{a}}: \mu<10 \end{array}

\newline

The sample size is

105

and the population standard deviation is assumed known with

\sigma=5

. Use

\alpha=0.05

\newline

(a) If the population mean is

9

, what is the probability that the sample mean leads to the conclusion do not reject

H_{0}

? (Round your answer to four decimal places.)

\newline

\square

\newline

(b) What type of error would be made if the actual population mean is

9

and we conclude that

H_{0}: \mu \geq 10

is true?

\newline

Type I

\newline

Type II

\newline

8

? (Round your answer to four decimal places. If it is not possible to commit a type II error enter NOT POSSIBLE.)

\newline

\square

Calculate z-score: For part (a), calculate the z-score for the sample mean when the population mean is $9$ using the formula $z = (\bar{x} - \mu) / (\sigma/\sqrt{n})$ , where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. $\newline$ Given: $\mu = 9$ , $\sigma = 5$ , $n = 105$ , and $\alpha = 0.05$ . $\newline$ The critical z-value for $\alpha = 0.05$ (one-tailed test to the left) can be found from the z-table or standard normal distribution table.
Find critical z-value: Find the critical z-value corresponding to $\alpha = 0.05$ . This is the z-value below which we would reject $H_0$ . For a left-tailed test, the critical z-value is $-1.645$ .
Calculate standard error: Calculate the standard error (SE) using the formula $SE = \frac{\sigma}{\sqrt{n}}$ . $SE = \frac{5}{\sqrt{105}} \approx 0.487$ .
Calculate z-score for sample mean: Calculate the z-score for the sample mean when $\mu = 9$ . $z = (9 - 10) / 0.487 \approx -2.053$ .
Find probability of not rejecting $H_0$ : Find the probability of not rejecting $H_0$ by looking up the z-score in the standard normal distribution table. The probability corresponding to $z = -2.053$ is $0.0202$ . However, since we want the probability of not rejecting $H_0$ , we need to find the complement of this probability. $P(\text{not reject } H_0) = 1 - 0.0202 = 0.9798$ . Round to four decimal places: $P(\text{not reject } H_0) = 0.9798$ .
Identify Type II error: For part (b), if the actual population mean is $9$ and we conclude that $H_0: \mu \geq 10$ is true, we would be making a Type II error because we failed to reject a false null hypothesis.
Calculate probability of Type II error: For part (c), calculate the probability of a Type II error when the actual population mean is $8$ . First, find the z-score for $\mu = 8$ using the same formula as before. $z = \frac{8 - 10}{0.487} \approx -4.106$ .
Find z-score for critical value: Look up the z-score of $-4.106$ in the standard normal distribution table. The probability corresponding to this z-score is very small, close to $0$ . However, the probability of a Type II error is the probability that the test statistic falls in the non-rejection region when the actual mean is $8$ . $P(\text{Type II error}) = P(z > -1.645 \text{ when } \mu = 8)$ .
Find z-score for critical value: Look up the z-score of $-4.106$ in the standard normal distribution table. The probability corresponding to this z-score is very small, close to $0$ . However, the probability of a Type II error is the probability that the test statistic falls in the non-rejection region when the actual mean is $8$ . $P(\text{Type II error}) = P(z > -1.645 \text{ when } \mu = 8)$ .To find $P(\text{Type II error})$ , we need to calculate the z-score for the critical value when $\mu = 8$ . $z = (\text{critical value} - 8) / 0.487$ . But we made a mistake; we should have used the z-score for the critical value when $\mu = 10$ to find the probability of not rejecting $H_0$ when the actual mean is $8$ .