We are given the following data:
Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22
We are to find the confidence intervals for 90%, 95% and 98% confidence level.
Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.
Part 1) 90% Confidence Intervalz value for 90% confidence interval = 1.645
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 90% confidence interval will be (110.69, 136.51)Part 2) 95% Confidence Intervalz value for 95% confidence interval = 1.96
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 95% confidence interval will be (108.22, 138.98)Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 98% confidence interval will be (105.34, 141.86)Part 4) Comparison of Confidence IntervalsThe 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)
As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
