Question

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

Answer 1

a. $134.66

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".

`\bar X =128` represent the sample mean for the sample

`\mu` 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:

`\bar X \pm z_(\alpha/2)(s)/(√(n))` (1)

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, 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
`z_(\alpha/2)=1.64`

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

`128-1.64(38)/(√(100))=121.768`

`128+1.64(38)/(√(100))=134.232`

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