Question

A recent graduate school study of a random sample of 250 US manufacturing companies determined the average financial report preparation time was 68.04 days with a standard deviation of 35.74 days. Calculate to three decimal places the 95 percent confidence interval for the mean report prep time for all US manufacturing companies. [63.001, 72.008] [63.957, 75.568] [63.505, 72.414] [61.612, 74.468] [63.612, 72.468]

Answer 1

`68.04-1.959(35.74)/(√(250))=63.618`

`68.04+1.959(35.74)/(√(250))=72.461`

And the best option for this case would be

Explanation:

Information given

`\bar X=68.04` represent the sample mean

`\mu` population mean

s=35.74 represent the sample standard deviation

n=250 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

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

The degrees of freedom are given by:

`df=n-1=250-1=249`

The Confidence level is 0.95 or 95%, and the significance
`\alpha=0.05` and
`\alpha/2 =0.025`, and the critical value for this case woud be
`t_(\alpha/2)=1.956`

And replacing we got:

`68.04-1.959(35.74)/(√(250))=63.618`

`68.04+1.959(35.74)/(√(250))=72.461`

And the best option for this case would be