In an effort to estimate the mean amount spent per household per month in a small Canadian town, data were collected for a sample of 50 households. The sample showed an average amount of $2000 and a standard deviation of $500.
Develop a 90% confidence interval estimate of the population mean amount spent.
Develop a 95% confidence interval estimate of the population mean amount spent.
Discuss what happens to the width of the confidence interval as the confidence level is increased. Does this seem reasonable? Explain.
If the data were collected for a sample of 20 households rather than 50, develop a 95% confidence interval estimate of the population mean amount spent, assuming the population has a normal probability distribution.
Discuss what happens to the width of the confidence interval as the sample size is decreased. Does this seem reasonable? Explain.
The formula would be: C. I. = M ± (z * SE)
z = the degree of alpha - we get it from a z-table: http://www.statsoft.com/textbook/sttable.html#z
SE = standard error
to get SE = standard deviation / square root of sample size
A. 90% confidence interval = our z score would be 1.65
our SE would be: 500/sq root 50
Confidence interval = 2000 ± (70.72* 1.65)
Lower level = 2000 - 116.68 = 1883.31
Upper level = 2000 + 116.68 = 2116.68
b) 95% confidence ...
This problem examines the math behind confidence intervals by looking at several mathematical cases. It gives a full explanation of the results.