Question

In a recent study, your town's chamber of commerce found that 73% of town residents believed that local businesses overcharged for their products. the source of this information was a telephone survey of 1,000 town residents. find a 95% confidence interval for the average percentage of everyone in your town who believes that local businesses overcharge.

Answer 1

The Confidence interval for 95% who believes that the local businesses overcharge = (0.7026, 0.7574)

What is meant by confidence interval?

The range of values we see in our sample and hope to identify the value that most closely represents the population are referred to as a confidence interval.

According to the given information:

Sample size n = 1000

73% of town residents believed that local businesses overcharged for their products over 1000 resident.

= (1000/100) x 73

= 730

Sample proportion p = 730/1000

                                    = .73

q = 1-p = 0.23

Std error of proportion = √(pq/n)

                                      = √((.73*0.27)/1000)

                                      = 0.0140

95% Z critical value = 1.96

Margin of error = 1.96*0.0140

                          = 0.0274

Confidence interval = sample proportion ±margin of error

(0.7026, 0.7574)

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