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A department store finds that in a random sample of 200 customers, 60% of the sampled customers had browsed its website prior to visiting the store. Based on this data, a 90% confidence interval for the population proportion of customers that browse the store’s website prior to visiting the store will be between

User Kobigurk
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Final answer:

To find a 90% confidence interval for the population proportion of customers that browse the store’s website prior to visiting the store, we can use the sample proportion and margin of error. The confidence interval is approximately 0.5506 to 0.6494.

Step-by-step explanation:

To find a 90% confidence interval for the population proportion of customers that browse the store’s website prior to visiting the store, we can use the formula:

Confidence Interval = Sample proportion ± Margin of Error

Given that 60% of the sampled customers had browsed the website, the sample proportion is 0.6. The margin of error can be calculated using the formula:

Margin of Error = Critical Value × Standard Error

For a 90% confidence level, the critical value can be found from a standard normal distribution table, which is approximately 1.645. The standard error can be calculated using the formula:

Standard Error = √((p × (1-p)) / n)

Substituting the values into the formulas, we can find:

Standard Error = √((0.6 × (1-0.6)) / 200) = 0.03

Margin of Error = 1.645 × 0.03 ≈ 0.0494

Finally, the confidence interval is:

Confidence Interval = 0.6 ± 0.0494 = (0.5506, 0.6494)

User Phu
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