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jake2920
04.04.2020 •
Mathematics
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 1291 tenth graders is drawn. Of the students sampled, 1098 read above the eighth grade level. Using the data, construct the 95% 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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Ответ:
The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.13, 0.168).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of
.
For this problem, we have that:
1291 tenth graders, 1098 read above the eighth grade level.
1291 - 1098 = 193 read at or below this level.
We want the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level.
So![n = 1291, \pi = \frac{193}{1291} = 0.149](/tpl/images/0582/0704/ec4c4.png)
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.13, 0.168).
Ответ:
480-18x=30 25 18-inch sections
Step-by-step explanation: