Question

A sample of eight workers in a clothing manufacturing company gave the following figures for the amount of time(in minutes) needed to join a collar to a shirt11 13 14 10 9 16 11 12Construct a 95% confidence interval for the true mean amount of time needed to join a collar.

Answer 1

`10.108 < \mu < 13.892`

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` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

We have the following distribution for the random variable:

`X \sim N(\mu , \sigma=0.45)`

And by the central theorem we know that the distribution for the sample mean is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

2) Confidence interval

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 mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=12`

The sample deviation calculated
`s=2.268`

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

`df=n-1=8-1=7`

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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that
`t_(\alpha/2)=2.36`

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

`12-2.36(2.268)/(√(8))=10.108`

`12+2.36(2.268)/(√(8))=13.892`

So on this case the 95% confidence interval would be given by (10.108;13.892)

`10.108 < \mu < 13.892`