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Maru [420]
3 years ago
6

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

ighth grade level. 1. Suppose a sample of 955 tenth graders is drawn. Of the students sampled, 812 read above the eighth grade level. Using the data, estimate the proportion of tenth graders reading at or below the eighth grade level. 2. Suppose a sample of 955 tenth graders is drawn. Of the students sampled, 812 read above the eighth grade level. Using the data, construct the 90% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level.
Mathematics
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

n= 955 represent the sampel size slected

x = 812 number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

The confidence interval for the proportion  would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the significance is \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution and we got.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

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this is an OER i dont know if u know what this is but I have explain the answer and how I got it :))​
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