Question

In order to estimate the average time spent on the computer terminals per student at local university, data were collected for a sample of 36 business students over a one-week period. The sample standard deviation is 1.8 hours. In order to construct a 90% t-based confidence interval for the population average time spent, the margin of error will be:

a) 0.1
b) 0.3
c) 0.4
d) 0.5

Answer 1

Answer:

Margin of Error = 0.5

Explanation:

Using this formula Z (s/√n)

Where

S = 1.8----------------------------- Standard Deviation

n = 36 ----------------------------------Number of observation

Z = 1.645 ------------------------------The chosen Z-value from the confidence table below

Confidence Interval || Z

80%. || 1.282

85% || 1.440

90%. || 1.645

95%. || 1.960

99%. || 2.576

99.5% || 2.807

99.9%. || 3.291

Substituting these values in the formula

Margin of Error = 1.645 * (1.8/√36)

Margin of Error = 1.645 * (1.8/6)

Margin of Error = 1.645 * 0.3

Margin of Error = 0.4935

Margin of Error = 0.5 (Approximated)