Question

Suppose we would like to gauge voters’ preferences for the President of Mars’ election. In an exit poll with a random sample of 120 voters, 30 voted for Mark Zuckerberg and 90 voted for Elon Musk. For the following, report your answers in 3 decimal places.

However in some applications, we would like to know whether Elon Musk is getting at least a certain percentage of the total votes. In this case, we should use a 100(1 − α)% lower one-sided confidence interval taking the form of [L, 1]. To construct a 100(1 − α)% lower one-sided Wald confidence interval through the pivot method, we start by setting P   X − p qp(1−p) n < zα   = 1 − α,

where X is the sample proportion, n the sample size and zα is the upper 100α% quantile of N(0, 1). Using this relationship, construct a 98% lower one-sided Wald confidence interval for p.

Answer 1

Here, n=120,

$p=\dfrac{90}{120}=0.75$

$\alpha=0.02$

$Z_{0.02}=2.326$

98% one sided lower confidence interval is-

$=p- Z_{0.02}\sqrt{\dfrac{p(1-p)}{n}}\\[9pt]=0.75-2.326\times \sqrt{\dfrac{0.75(1-0.75)}{120}}\\[9pt]=0.75-2.326\times 0.03952\\[9pt]=0.75- 0.09194\\[9pt]= 0.6580$