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Moving to another question will save this response. «| || Question 11 of 26 > » Question 11 0.04 points Save Answer A student organization wants to start a nightclub for students under the age of 21. They have contacted an agency that is willing to fund their proposal with the one condition that the proportion of students who respond favorably to the proposal is at least 75%. To assess support for the proposal, they collect a random sample of 100 students and ask each respondent if he or she would patronize with this type of establishment and find that 82 respondents respond favorably. They conduct a huypothesis test to determine if the sample evidence allow them to receive the funding from the agency. Using significance level of 1% the correct conclusion is: O Reject Ho, there is sufficient evidence that the proportion of students who favor the proposal is at least 75% O Do Not Reject Ho, there is sufficient evidence that the proportion of students who favor the proposal is at least 75% O Reject Ho, there is insufficient evidence that the proportion of students who favor the proposal is at least 75% O Do Not Reject Ho, there is insufficient evidence that the proportion of students who favor the proposal is at least 75% Moving to another question will save this response. Question 11 of 26

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11)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P > 0.75
Alternative hypothesis: P < 0.75

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.0433
z = (p - P) /

z = 1.62

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 1.62.

Thus, the P-value = 0.0526

Interpret results. Since the P-value (0.0526) is greater than the significance level (0.01), we have to accept the null hypothesis.

Do not reject H0, there is sufficient evidence that the proportion of students who favors the proposal is atleast 75%.

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