Question

A survey of 225 students showed the mean number of hours spent studying per week was 20.6 and the standard deviations was 2.7

Answer 1

.3 is the answer.

Explanation:

Answer 2

The margin of error is approximately 0.3

Explanation:

The following information has been provided;

The sample size, n =225 students

The sample mean number of hours spent studying per week = 20.6

The standard deviation = 2.7

The question requires us to determine the margin of error that would be associated with a 90% confidence level. In constructing confidence intervals of the population mean, the margin of error is defined as;

The product of the associated z-score and the standard error of the sample mean. The standard error of the sample mean is calculated as;

`(sigma)/(√(n) )`

where sigma is the standard deviation and n the sample size. The z-score associated with a 90% confidence level, from the given table, is 1.645.

The margin of error is thus;

`1.645*(2.7)/(√(225))=0.2961`

Therefore, the margin of error is approximately 0.3