Question

In a time-use study, 20 randomly selected college students were found to spend a mean of 1.4 hours on the internet each day. The standard deviation of the 20 scores was 1.3 hours.
Construct a 90% confidence interval for the mean time spent on the internet by college students. Assume t* = 1.729
If you increased the sample size for the problem above, what would happen to your confidence interval, assuming the sample mean and standard deviation remain unchanged?

Answer 1

Solution :

Given that,

= 1.4

s = 1.3

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,19 = 1.729

Margin of error = E = t_/2,df * (s /n)

= 1.729 * (1.3 / 20)

= 0.5

The 90% confidence interval estimate of the population mean is,

- E < < + E

1.4 - 0.5< < 1.4 + 0.5

0.9 < < 1.9

(0.9 , 1.9)

If you increased the sample size for the problem above confidence interval, the sample mean and

standard deviation remain unchanged .

but standard errr is changed .