Question

Suppose a sample of 2404 tenth graders is drawn. Of the students sampled, 1803 read above the eighth grade level. Using the data, construct the 98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level. Round your answers to three decimal places.

Answer 1

98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is [0.229 , 0.271].

Explanation:

We are given that a sample of 2404 tenth graders is drawn. Of the students sampled, 1803 read above the eighth grade level.

Firstly, the pivotal quantity for 98% confidence interval for the population proportion is given by;

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

where,
`\hat p` = proportion of tenth graders reading at or below the eighth grade level in a sample of 2404 =
`1 - (1803)/(2404)` = 0.25 {because in question we are the no. of students who read above the eighth grade level}

n = sample of tenth graders = 2404

p = population proportion

Here for constructing 98% confidence interval we have used One-sample z proportion statistics.

So, 98% confidence interval for the population proportion, p is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98 {As the critical value of z at 1%

significance level are -2.3263 & 2.3263}

P(-2.3263 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 2.3263) = 0.98

P(
`-2.3263 * {\sqrt{(\hat p(1-\hat p))/(n) } }` <
`{\hat p-p}` <
`2.3263 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.98

P(
`\hat p-2.3263 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+2.3263 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.98

98% confidence interval for p =[
`\hat p-2.3263 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ,
`\hat p+2.3263 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ]

= [
`0.25-2.3263 * {\sqrt{(0.25(1-0.25))/(2404) } }` ,
`0.25+2.3263 * {\sqrt{(0.25(1-0.25))/(2404) } }` ]

= [0.229 , 0.271]

Therefore, 98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is [0.229 , 0.271].

Answer 2

(0.229; 0.271)

Explanation:

Sample size (n) =2404

Z-score for a 98% confidence interval (z) = 2.33

The proportion 'p' of students reading at or below the eighth grade level is:

`p=1-(1803)/(2404)\\p=0.25`

The confidence interval, assuming a normal distribution, is given by:

`p \pm z\sqrt{(p*(1-p))/(n)}`

Applying the given data, the lower (L) and upper (U) bounds of the interval are:

`L=0.25-2.33*\sqrt{(0.25*(1-0.25))/(2404)} \\L=0.229\\U=0.25+2.33*\sqrt{(0.25*(1-0.25))/(2404)} \\U=0.271`

The confidence interval is I = (0.229; 0.271)