You are conducting a study to see if the probability of catching the flu this ye

Question

You are conducting a study to see if the probability of catching the flu this year is significantly more than 0.31. You use a significance level of alpha = 0.002. H_0: p = 0.31 H_1: p > 0.31 You obtain a sample of size n = 307 in which there are 102 successes. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is.. less than (or equal to) alpha greater than alpha This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the probability of catching the flu this year is more than 0.31. There is not sufficient evidence to warrant rejection of the claim that the probability of catching the flu this year is more than 0.31. The sample data support the claim that the probability of catching the flu this year is more than 0.31. There is not sufficient sample evidence to support the claim that the probability of catching the flu this year is more than 0.31.

Explanation / Answer

Solution:-

x = 102

n = 307

p = 102/307

p = 0.3322

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

Null hypothesis: P < 0.31

Alternative hypothesis: P > 0.31

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 0.002. 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 ]

= 0.0264

z = (p - P) /

a) z = 0.843

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 0.843.

b) Thus, the P-value = 0.2005

c) Interpret results. Since the P-value (0.20) is greater than the significance level (0.002), we cannot accept the null hypothesis.

d) We fail to reject the null hypothesis.

e) There is sufficient evidence to warrant the rejection of the claim that the probability of catching the flu this year is more than 0.31.

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