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A random sample of n = 1000 voters is taken and 650 of those surveyed indicate...

A random sample of n = 1000 voters is taken and 650 of those surveyed indicate that they like pizza. What is the value of the sample proportion of those who like pizza? Provide a 95% confidence interval estimate of the proportion of individuals in the population who like pizza. Interpret your results.

Math   Statistics-And-Probability   
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Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 650 / 1000 = 0.650

1 - = 1 - 0.650 = 0.35

Z/2 = 1.96

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 * (((0.650 * 0.35) / 1000)

Margin of error = E = 0.030

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.650 - 0.030 < p < 0.650 + 0.030

0.620 < p < 0.680

The 95% confidence interval for the population proportion p is : 0.620 , 0.680

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