Question

Suppose that the number of people employed by an individual business is normally distributed with an unknown mean and standard deviation. A random sample of 17 businesses is taken and gives a sample mean of 15 employees and a sample standard deviation of 2 employees. df to.10 to.025 to.o1 to.o05 to.05 1.341 1.753 2.131 2.602 2.947 15 1.337 1.746 2.120 2.583 2.921 16 17 1.333 1.740 2.110 2.567 2.898 1.330 1.734 2.101 2.552 2.878 18 19 1.328 1.729 2.093 2.539 2.861 Find the margin of error, for a 95% confidence interval estimate for the population mean using the Student's t-distribution. Use the portion of the table above. Round the final answer to two decimal places.

Answer 1

The interval is (13.97, 16,03).

Explanation:

The formula for estimating a popoulation mean through a confidence interval using Student's t-distribution is given by:

`\bar{x}\pm t_(n-1) * s/√(n)`

Where:

`\bar{x}` is the sample's mean,

`t_(n-1)` is t-score with n-1 degrees of freedom,

`s` is the standard error,

`n` is the sample's size.

We have that:

`\bar{x}=15`

`t_(n-1)=2.12` (found in the table which i've attached)

`s=2`

`n=17`

Now we can replace in the formula to obtain the confidence interval:

`\bar{x}\pm t_(n-1) * s/√(n)=15\pm 2.12 * 2/√(17)= (13.97, 16,03)`

Therefore we can conclude that the population's mean is estimated to be with 95% confidence within the interval (13.97, 16,03).