Question

A large company wants to estimate the average amount of money that is owed to it by delinquent customers, µ. The mean of this sample is $230 and the standard deviation is $50. Construct a 95% and 99% confidence interval for the population parameter µ.

This is the only information given. There is no sample size given.

Answer 1

Then for solving we have to assume the sample size n = 100

Construct a 95% confidence interval for the population parameter:

n=100

Mean =230

SD=50

95% Table value t at 100-1 = 99 DF =1.9842

$CI = \bar x \pm t* \frac {s}{\sqrt{n}}$

$= 230 \pm 1.9842* \frac {50}{\sqrt{100}}$

= (220.08, 239.92)

B). Construct a 99% confidence interval for the population parameter (μ)

99% Table value t at 100-1 = 99 DF =2.6264

$CI = \bar x \pm t* \frac {s}{\sqrt{n}}$

$= 230 \pm 2.6264* \frac {50}{\sqrt{100}}$

= (216.87, 243.13)