The mean monthly expenditure on gasoline per household in middletown is determined by selecting a random sample of 100 households. the sample mean is $128, with a sample standard deviation of $38, what is the upper bound of a 90% confidence interval for the mean monthly expenditure on gasoline per household in middletown? a. $134.66 b. $162.20 c. $129.38 d. $131.42 e. $13287

a. $134.66

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean (variable of interest)

s=38 represent the sample standard deviation

n=100 represent the sample size  

2) Calculate the confidence interval

Since the sample size is large enough n>30. The confidence interval for the mean is given by the following formula:

  (1)

Since the Confidence is 0.90 or 90%, the value of and , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that

Now we have everything in order to replace into formula (1):

The closest value would be $134.66 and that would be the answer for this case.

answer is more than one

step-by-step explanation:

any options?

step-by-step explanation: