Question

⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football.

Answer 1

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

P.Q. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults =
`(354)/(1219)`

n = sample of US adults = 1291

p = population proportion of adults

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

significance level are -1.96 & 1.96}

P(-1.96 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 1.96) = 0.95

P(
`-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) }` <
`{\hat p-p}` <
`1.96 * {\sqrt{(\hat p(1-\hat p))/(n) }` ) = 0.95

P(
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) }` < p <
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) }` ) = 0.95

95% confidence interval for p = [
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) }` ,
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) }` ]

= [
`(354)/(1219)-1.96 * {\sqrt{((354)/(1219)(1-(354)/(1219)))/(1219) }` ,
`(354)/(1219)+1.96 * {\sqrt{((354)/(1219)(1-(354)/(1219)))/(1219) }` ]

= [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].