Question

An Office of Admissions document claims that 54.6% of UVA undergraduates are female. To test whether this claim is accurate, a random sample of 225 UVA undergraduates was selected. In this sample, 54.6667% were female. Is there sufficient evidence to conclude that the document's claim is false? Carry out a hypothesis test at a 1% significance level.

The p-value is:

Answer 1

Solution

P = 0.546

n = 225

p = 0.546667

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

Null hypothesis: P = 0.546

Alternative hypothesis: P ≠ 0.546

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 (σ) and compute the z-score test statistic (z).

σ = sqrt[ P * ( 1 - P ) / n ]

=sqrt(0.546*(1-0.546)/225)

σ = 0.033192

z = (p - P) / σ

= (0.546667-0.546)/0.33192

z = 0.00201

z_critical = + 2.575

Rejection region:-

- 2.575 > z > 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 - 0.00201 or greater than 0.00201.

Thus, the P-value = 0.0016

Interpret results. Since the P-value (0.0016) is lesser than the significance level (0.01), we can reject the null hypothesis.