Question

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Out of a sample of 500 high school students, 376 said they would prefer to have computers in every classroom. Construct a 95% confidence interval for the population mean of high school students who would prefer to have computers in every classroom.

CI = (71.41%, 78.99%)
CI = (71.98%, 79.45%)
CI = (70.23%, 80.17%)
CI = (72.02%, 78.38%)

Answer 1

Answer:

CI = (71.41%, 78.99%)

Step-by-step explanation:

$\frac{376}{500} - 1.96 \sqrt{ \frac{ (\frac{376}{500} )( \frac{124}{500} )}{500} } = 71.415\%$

$\frac{376}{500} + 1.96 \sqrt{ \frac{ (\frac{376}{500} )( \frac{124}{500}) }{500} } = 78.985\%$

Answer 2

If 500 high school students, 376 said they would prefer to have computers in every classroom then the confidence interval at 95% is (71.41%,78.99%)  Since option (a) is correct.

Given x= 376, n = 500

The (71.41%,78.99%)  

$\hat{p}=\frac{x}{n}\\ =\frac{376}{500} \\=0.752$

Critical values

Using the z-table the z-critical value at the given significance level at α=0.05 is,$z_{a/2} =1.96$

Construct a 95% confidence interval

That is,

$\hat{p}= z_{a/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n} }$

  $=0.752\pm 1.96\sqrt{\frac{0.752(1-0.752)}{500} }$

  $=0.752\pm 0.0379$

  = (71.41%,78.99%)  

Learn more about confidence intervals here: https://brainly.com/question/15712887

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