Question

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level. Step 2 of 2: Suppose a sample of 2089 tenth graders is drawn. Of the students sampled, 355 read at or below 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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Answer #1

Solution :

Given that,

n = 2085

x = 355

Point estimate = sample proportion = = x / n = 355/2089=0.170

1 -   = 1-0.170 =0.83

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z / 2 *( (( * (1 - )) / n)

= 2.326 (((0.170*0.83) /2085 )

E = 0.019

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.170-0.019 < p <0.170+ 0.019

0.151< p < 0.189

The 98% confidence interval for the population proportion p is : 0.151< p < 0.189

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