Answer:
The margin of error is approximately 0.3
Step-by-step 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;
[tex]\frac{sigma}{\sqrt{n} }[/tex]
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;
[tex]1.645*\frac{2.7}{\sqrt{225}}=0.2961[/tex]
Therefore, the margin of error is approximately 0.3
