Mathematics, 30.03.2020 23:40, mrylenastewart

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.

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answered: stormserena

(0.229; 0.271)

Step-by-step 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:

$p=1-\frac{1803}{2404}\\p=0.25$

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

$p \pm z\sqrt{\frac{p*(1-p)}{n}}$

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

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

The confidence interval is I = (0.229; 0.271)

answered: diamond2738

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

Step-by-step 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. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of tenth graders reading at or below the eighth grade level in a sample of 2404 = $1 - \frac{1803}{2404}$ = 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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.3263) = 0.98

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

P( $\hat p-2.3263 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.3263 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.98

98% confidence interval for p =[ $\hat p-2.3263 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.3263 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.25-2.3263 \times {\sqrt{\frac{0.25(1-0.25)}{2404} } }$ , $0.25+2.3263 \times {\sqrt{\frac{0.25(1-0.25)}{2404} } }$ ]