Question

a local company has 93 insurance salespersons. the boss wants to know if the salespersons would prefer a weekend meeting instead of a meeting during the week. she asked 24 of the salespersons if they would prefer a weekend meeting and 15 of them said that they would prefer a weekend meeting. (a) find the point estimate of the population proportion that would prefer a weekend meeting. (round your answer to 3 decimals.) (b) calculate the margin of error. (round your answer to 3 decimals.) (c) calculate the 99 percent confidence interval of the population proportion that would prefer a weekend meeting. (round each number to 3 decimals.) [ , ]

Answer 1

(a) The point estimate of the population proportion that would prefer a weekend meeting is 0.625.

This can be calculated by dividing the number of salespersons that would prefer a weekend meeting (15) by the total number of salespersons surveyed (24).

The margin of error is 0.094. This can be calculated by multiplying the standard error of the sample proportion (0.039) by the critical value (2.326).

(c) The 99% confidence interval is [0.438, 0.813]. This can be calculated by subtracting and adding the margin of error to the point estimate. The lower bound is calculated by subtracting the margin of error from the point estimate and the upper bound is calculated by adding the margin of error to the point estimate.

In conclusion, the point estimate of the population proportion that would prefer a weekend meeting is 0.625, the margin of error is 0.094 and the 99% confidence interval is [0.438, 0.813].

For more questions like Point estimate click the link below:

https://brainly.com/question/28273976

#SPJ4