You are conducting a study to see if the proportion of women over 40 who regular

Question

You are conducting a study to see if the proportion of women over 40 who regularly have mammograms is significantly more than 0.9. You use a significance level of =0.002. H0:p=0.9 H1:p>0.9 You obtain a sample of size n=547 in which there are 515 successes.

What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value =

The p-value is... (choose 1)

less than (or equal to)

greater than

This p-value leads to a decision to... (choose 1)

reject the null

accept the null

fail to reject the null

As such, the final conclusion is that... (choose 1)

There is sufficient evidence to warrant rejection of the claim that the proportion of women over 40 who regularly have mammograms is more than 0.9.

There is not sufficient evidence to warrant rejection of the claim that the proportion of women over 40 who regularly have mammograms is more than 0.9.

The sample data support the claim that the proportion of women over 40 who regularly have mammograms is more than 0.9.

There is not sufficient sample evidence to support the claim that the proportion of women over 40 who regularly have mammograms is more than 0.9.

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

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

Null hypothesis: P = 0.90
Alternative hypothesis: P > 0.90

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

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 ] = sqrt [(0.9 * 0.1) / 547] = 0.01282707374
z = (p - P) / = (0.94 - 0.90)/0.01282707374 = 3.1184

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 more than 3.1184. We use the Normal Distribution Calculator to find P(z > 3.1184).

The P-Value is 0.000909.
The result is significant at p < 0.05.

Interpret results. Since the P-value (0.000909) is less than the significance level (0.002), we cannot accept the null hypothesis.

Conclusion. The sample data support the claim that the proportion of women over 40 who regularly have mammograms is more than 0.9.

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