Question

The University College is interested in the average number of hours per week that freshmen spend going to parties. They took a random sample of 300 freshmen and calculated a mean of 6 hours a week going to parties and a 95% confidence interval for the population mean to be 4 to 8 hours a week. Based on this information, what was their standard error of the mean?

Group of answer choices

4

2

1

.67

Answer #1

Solution :

Sample mean = = (4 + 8) / 2 = 6

Margin of error = E = Upper confidence interval - = 8 - 6 = 2

E = Z_{/2}*
(
/n)

(
/n)
= Z_{/2}
/ E = 1.96 / 2 = 0.98 = 1

standard error of the mean = **1**

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