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Georgia [21]
3 years ago
15

A survey of an urban university (population of 25,450) showed that 883 of 1,112 students sampled supported a fee increase to fun

d improvements to the student recreation center. Using the 95% level of confidence, what is the confidence interval for the proportion of students supporting the fee increase
Mathematics
1 answer:
quester [9]3 years ago
6 0

Answer:

The confidence interval for the proportion of students supporting the fee increase

( 0.77024, 0.81776)

Step-by-step explanation:

<u>Explanation:</u>

Given data a survey of an urban university (population of 25,450) showed that 883 of 1,112 students sampled supported a fee increase to fund improvements to the student recreation center.

Given sample size 'n' = 1112

Sample proportion 'p' = \frac{883}{1112} = 0.7940

                           q = 1 - p = 1- 0.7940 = 0.206

<u>The 95% level of confidence intervals</u>

The confidence interval for the proportion of students supporting the fee increase

(p-z_{\alpha } \sqrt{\frac{pq}{n} } ,p + z_{\alpha } \sqrt{\frac{pq}{n} } )

The Z-score at 95% level of significance =1.96

(0.7940-1.96\sqrt{\frac{0.7940 X 0.206}{1112} } ,0.7940 + 1.96 \sqrt{\frac{0.7940 X 0.206}{1112} } )

(0.7940-0.02376 , 0.7940+0.02376)

( 0.77024, 0.81776)

<u>Conclusion:</u>-

The confidence interval for the proportion of students supporting the fee increase

( 0.77024, 0.81776)

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