Home page for accesible maths 4.1 Inference for a single proportion 4.1.2 Confidence intervals for a proportion 4.2 Difference of two proportions

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4.1.3 Hypothesis testing for a proportion

To apply the normal distribution framework in the context of a hypothesis test for a proportion, the independence and success-failure conditions must be satisfied. In a hypothesis test, the success-failure condition is checked using the null proportion: we verify $np_{0}$ and $n(1-p_{0})$ are at least 10, where $p_{0}$ is the null value.

Example 4.1.3

Deborah Toohey is running for Congress, and her campaign manager claims she has more than 50% support from the district’s electorate. Set up a one-sided hypothesis test to evaluate this claim.

Answer. Is there convincing evidence that the campaign manager is correct? $H_{0}:p=0.50$ , $H_{A}:p>0.50$ .

Example 4.1.4

A newspaper collects a simple random sample of 500 likely voters in the district and estimates Toohey’s support to be 52%. Does this provide convincing evidence for the claim of Toohey’s manager at the 5% significance level?

Answer. Because this is a simple random sample that includes fewer than 10% of the population, the observations are independent. In a one-proportion hypothesis test, the success-failure condition is checked using the null proportion, $p_{0}=0.5$ : $np_{0}=n(1-p_{0})=500\times 0.5=250>10$ . With these conditions verified, the normal model may be applied to $\hat{p}$ .

Next the standard error can be computed. Answer. The null value is used again here, because this is a hypothesis test for a single proportion.

SE=\sqrt{\frac{p_{0}\times(1-p_{0})}{n}}=\sqrt{\frac{0.5\times(1-0.5)}{500}}=0% .022

A picture of the normal model is shown in Figure LABEL:pValueForCampaignManagerClaimOfMoreThan50PercentSupport with the p-value represented by the shaded region. Based on the normal model, the test statistic can be computed as the Z score of the point estimate:

Z=\frac{\text{point estimate}-\text{null value}}{SE}=\frac{0.52-0.50}{0.022}=0% .89

The upper tail area, representing the p-value, is $1-\mathbb{P}(Z<0.89)=1-0.8132671=0.1867$ . Because the p-value is larger than 0.05, we do not reject the null hypothesis, and we do not find convincing evidence to support the campaign manager’s claim.

Hypothesis test for a proportion Set up hypotheses and verify the conditions using the null value, $p_{0}$ , to ensure $\hat{p}$ is nearly normal under $H_{0}$ . If the conditions hold, construct the standard error, again using $p_{0}$ , and show the p-value in a drawing. Lastly, compute the p-value and evaluate the hypotheses.