Question

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 28, 23, 18, 15, 16, 5, 21, 13. [You may find it useful to reference the t table.]

a. Calculate the sample mean and the sample standard deviation. (Round intermediate calculations to at least 4 decimal places. Round "Sample mean" to 3 decimal places and "Sample standard deviation" to 2 decimal places.)

b. Construct the 90% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

c. Construct the 95% confidence interval for the population mean. (Round "t" value to 3 decimal places and final answers to 2 decimal places.)

d. What happens to the margin of error as the confidence level increases from 90% to 95%?

Answer 1

a) Since we know that

and n = 8

This implies that

b)

Confidence interval(in %) = 90

t = 1.8946

Since we know that

Required confidence interval = (17.375-4.6554, 17.375+4.6554)

Required confidence interval = (12.7196, 22.0304)

c)

Confidence interval(in %) = 95

t = 2.3646

Since we know that

Required confidence interval = (17.375-5.8103, 17.375+5.8103)

Required confidence interval = (11.5647, 23.1853)

d) when we increase the confidence interval, t value increases and so the value of margin of error is increased