An Office of Admissions document claims that 55.6% of UVA undergraduates are fem

Question

An Office of Admissions document claims that 55.6% of UVA undergraduates are female. To test this claim, a random sample of 240 UVA undergraduates was selected. In this sample, 54.5833% were female. Is there sufficient evidence to conclude that the document's claim is false? Carry out a hypothesis test at a 2% significance level. hint: We are testing whether the Office of Admissions’ claim is accurate (neither too high nor too low). So, we would be interested in a sample result that was either much higher or much lower than 55.6%.

A. The p-value is

I got .7490 but is was marked wrong

Explanation / Answer

Solution:-

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

Null hypothesis: P = 0.556
Alternative hypothesis: P 0.556

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 0.02. 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 = - 0.32

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 - 0.32 or greater than 0.32.

Thus, the P-value = 0.749

Interpret results. Since the P-value (0.749) is greater than the significance level (0.02), we cannot reject the null hypothesis.

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