Question

In a survey of 447 registered voters, 157 of them wished to see Mayor Waffleskate lose her next election. The Waffleskate campaign claims that no more than 27% of registered voters wish to see her defeated. Does the 98% confidence interval for the proportion support this claim? (Hint: you should first construct the 98% confidence interval for the proportion of registered voters who wish to see Waffleskate defeated.) (0.299, 0.404)
A. No

B. Yes

Answer 1

Answer with explanation:

Let p be the true proportion of  registered voters wish to see Mayor Waffleskate defeated.

As per given , we have

$H_0: p\leq0.27\\\\ H_a: p >0.27$

Sample size : n= 447

Number of of registered voters wish to see Mayor Waffleskate defeated = 157

I.e. sample proportion :  $\hat{p}=\dfrac{157}{447}\approx0.3512$

Confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where n= sample size

$\hat{p}$ = sample proportion

z* = critical z-value.

Critical z-value for 98% confidence interval is 2.33.  (By z-table)

Then, the 98% confidence interval for the proportion of registered voters who wish to see Waffleskate defeated will be :

$0.3512\pm2.33\sqrt{\dfrac{0.3512(1-0.3512)}{447}}\\\\=0.3512\pm (2.33)(0.022577656)\\\\=0.3512\pm 0.05260593848\\\\=(0.3512-0.05260593848,\ 0.3512+0.05260593848)\\\\=(0.29859406152,\ 0.40380593848)\approx(0.299,\ 0.404)$

Since the 0.27 < 0.299 , it means 0.27 does not belong to the above confidence interval.

So , we reject the null hypothesis ($H_0$).

So ,  98% confidence interval does not support the claim.