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.
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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