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A national survey of insurance offices was taken, resulting in a random sample of 244 companies. Of
these 244 companies, 196 responded that they were going to purchase new software for their offices in the next year. Construct a
90% confidence interval to estimate the population proportion of insurance offices that intend to purchase new software during the
next year.
Appendix A Statistical Tables
(Round the Intermediate values to 2 decimal places, e.g. 0.25. Round your answers to 3 decimal places, e.g. 0.253.)
sps

Answer 1

To estimate the population proportion of insurance offices that plan to purchase new software, a 90% confidence interval was constructed using the sample proportion and the z-value corresponding to the 90% confidence level. The resulting interval, after rounding to three decimal places, is (0.762, 0.844).

Step-by-step explanation:

To construct a 90% confidence interval to estimate the population proportion of insurance offices that intend to purchase new software during the next year, we can use the formula for the confidence interval for a population proportion:

CI = p± (z* ∙ √(p(1-p)/n))

Where:

• p is the sample proportion = 196/244
• n is the sample size = 244
• z* is the z-value for the 90% confidence level (from a z-table, approximately 1.645 for a 90% confidence interval)

First, calculate the sample proportion p:

p = 196/244 ≈ 0.803

Now, calculate the standard error (SE) of the sample proportion:

SE = √(p(1-p)/n)

SE = √(0.803(1-0.803)/244) = √(0.803(0.197)/244) ≈ √(0.158041/244) ≈ √(0.0006479) ≈ 0.025

Next, calculate the margin of error (ME):

ME = z* ∙ SE

ME = 1.645 ∙ 0.025 ≈ 0.041

Finally, calculate the confidence interval:

CI = 0.803 ± 0.041

The 90% confidence interval is (0.762, 0.844).

Rounding the values to three decimal places, the 90% confidence interval is (0.762, 0.844).