Question

1. Suppose the true proportion of voting Seattleites who support the $15 per hour wage increase is to be estimated. If we took a simple random sample of 805 likely voters in Seattle, and calculated the sample proportion who plan to vote in support of the proposition is 65%, then:a) what would be the standard error of the statistic?b) Calculate a 98% confidence interval for the proportion of voters in Seattle in support. Interpret.

Answer 1

Let $\hat{p}$ be the sample proportion.

When n is large and p is not close to zero or one, we can use the normal distribution to approximate the binomial.

The sampling distribution of $\hat{p}$ is normal.

The mean of the sampling distribution is $p.$

The standard deviation of the sampling distribution (called the standard error) is

$$SE=\sqrt{{p(1-p) \over n}}$$

Given that $p=0.65, n=805$

a)

$$SE=\sqrt{{0.65(1-0.65) \over 805}}\approx0.016811$$

b)

Confidence Level 98%: $z^*-$ value$=2.33$

Confidence interval for $p$ is

$$\hat{p}\pm z^*\times SE$$

$$0.65\pm 2.33\times 0.016811\approx0.65\pm0.039$$

$$98\%\ CI=(0.611, 0.689)$$