A firm employs 189 junior accountants. In a random sample of...

A firm employs 189 junior accountants. In a random sample of 50 of these, the mean number of hours overtime billed in a particular week was 9.7, and the sample standard deviation was 6.2 hours. a. Find a $95 \%$ confidence interval for the mean number of hours overtime billed per junior accountant in this firm that week. b. Find a $99 \%$ confidence interval for the total number of hours overtime billed by junior accountants in the firm during the week of interest.

A firm employs 189 junior accountants. In a random sample of 50 of these, the mean number of hours overtime billed in a particular week was 9.7, and the sample standard deviation was 6.2 hours.
a. Find a $95 \%$ confidence interval for the mean number of hours overtime billed per junior accountant in this firm that week.
b. Find a $99 \%$ confidence interval for the total number of hours overtime billed by junior accountants in the firm during the week of interest.

Key Concepts

Finite Population Correction

In instances where sampling is done without replacement from a finite population, the finite population correction (FPC) factor is applied to adjust the standard error. This correction reduces the variability estimate because the sample constitutes a non-negligible portion of the entire population.

Transformation of Confidence Interval for Totals

When the parameter of interest involves a total sum rather than an average, the confidence interval for the mean can be scaled appropriately by multiplying the endpoints by the total number of units in the population. This transformation maintains the confidence level while providing an interval estimate for the aggregate quantity.

Confidence Interval

A confidence interval provides a range of values that is likely to contain the true population parameter, such as a mean or total, with a specified level of confidence. It incorporates the point estimate, the variability in the data, and the critical value from the appropriate sampling distribution to reflect uncertainty due to sampling.

t-Distribution

When the population standard deviation is unknown and the sample size is not exceedingly large, the t-distribution is used instead of the normal distribution. The t-distribution accounts for the additional uncertainty introduced by estimating the standard deviation from the sample and is characterized by degrees of freedom.

Transcript

00:01 So we want to find in part a a 95 % confidence interval for the population mean.

00:08 And we have the population size is 189, and they have sampled 50 people and found that the mean billing overtime hours was 9 .7 with a sample standard deviation of 6 .2.

00:23 And so we can see that this is more than 5 % of the population.

00:29 So we have to use.

00:30 That correction factor.

00:32 So we're going to have 9 .7 plus or minus and now we have our t value with 49 degrees of freedom and having 0 .025 in the upper tail.

00:43 That's corresponding to 2 .010.

00:47 And so we'll have that 2 .010.

00:50 With a normal distribution, it would be 1 .96.

00:53 So it's a little bit bigger than that.

00:55 And next we need to multiply by the tradition.

01:00 Standard error, which would end up being that 6 .2 divided by the square root of n, but then we need to multiply it by that correction factor of 189 minus the sample size, and then the 189 minus 1, so 188.

01:20 And so when we put that in, we have the 9 .7 minus 2 .01 times 6 .2 divided by the square, root of 50 and because we don't have the population size is not large enough compared to that sample size we need to now multiply it by that that would be 139 divided by 188 a square rooted and so when we do that calculation we get 8 .188 and it comes out for six if you need more decimal places but we'll just use one more than what was given to us for the mean.

02:04 And then the upper limit, upper confidence limit, will be changing that to an addition sign, and that comes out to be 11 .215...

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