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A software company allows their programmers to work from home an average of 20 hours per week. A department of 15 programmers work a mean of 22 hours per week from home with a standard deviation of 5.6 hours. At is their sufficient evidence to conclude a difference? Give a 95% confidence interval for the true mean number of hours working from home

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Answer:

The 95% confidence interval would be given by (18.906;25.094)

Since the confidence interval contains the value 20, we don't have enough evidence to conclude a difference.

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 =22 represent the sample mean for the sample


\mu population mean (variable of interest)

s=5.6 represent the sample standard deviation

n=15 represent the sample size

2) Solution to the problem

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


\bar X \pm t_(\alpha/2)(s)/(√(n)) (1)

In order to calculate the critical value
t_(\alpha/2) we need to find first the degrees of freedom, given by:


df=n-1=15-1=14

Since the Confidence is 0.95 or 95%, the value of
\alpha=0.05 and
\alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,14)".And we see that
t_(\alpha/2)=2.14

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


22-2.14(5.6)/(√(15))=18.906


22+2.14(5.6)/(√(15))=25.094

So on this case the 95% confidence interval would be given by (18.906;25.094)

Since the confidence interval contains the value 20, we don't have enough evidence to conclude a difference.

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