Question

A random sample of 15 group leaders, supervisors, and similar personnel revealed that a person spent an average 6.5 years on the job before being promoted. The standard deviation of the sample was 1.7 years. Using the 0.80 degree of confidence, what t-value should be used in computing an interval estimate

Answer 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 0.8 or 80%, the value of
`\alpha=0.2` and
`\alpha/2 =0.1`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.1,14)".And we see that
`t_(\alpha/2)=1.345`

Explanation:

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=6.5` represent the sample mean

`\mu` population mean (variable of interest)

s=1.7 represent the sample standard deviation

n=15 represent the sample size

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 0.8 or 80%, the value of
`\alpha=0.2` and
`\alpha/2 =0.1`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.1,14)".And we see that
`t_(\alpha/2)=1.345`

Answer 2

T = 1.345

Explanation:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 15 - 1 = 14

80% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.8)/(2) = 0.9([tex]t_(9)`). So we have T = 1.345