Decide whether the normal sampling distribution can be used. Ifit can be used, t

Question

Decide whether the normal sampling distribution can be used. Ifit can be used, test the claim about the population proportion p at the given level of significance using the given sample statistics Claim, p #029; = 0.01; Sample statistics: p-0.26, n-200 Can the normal sampling distribution be used? A. Yes, because both np and nq are greater than or equal to 5 O B. No, because ng is less than 5. C. Yes, because pq is greater than -0.01. D. No, because np is less than 5. State the null and altenative hypotheses. OA. Ho:p 20.29 H p 0.29 (y B. Ho:p=0.29 Ha: p #0.29 C. Ho:ps0.29 Ha p> 0.29 D. The test cannot be performed

Explanation / Answer

Solution:-

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

Null hypothesis: P = 0.29

Alternative hypothesis: P 0.29

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it 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 ]

= 0.0321

z = (p - P) /

z = - 0.935

zcritical = 2.575

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 two-tailed test, the P-value is the probability that the z-score is less than -0.935 or greater than 0.935.

Thus, the P-value = 0.348

Interpret results. Since the P-value (0.348) is greater than the significance level (0.01), we cannot reject the null hypothesis.

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