Question

A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a standard deviation of 2 years. At 95% confidence, test to determine if the average age of all the students at the university is significantly more than 24.

Answer 1

`t=(25-24)/((2)/(√(16)))=2`

The degrees of freedom are:

`df = n--1=16-1=15`

Now since we are conducting a right tailed test the p value can be calculated as follows:

`p_v =P(t_(15)>2)=0.0320`

Since the p value is lower than the significance level of
`\alpha=0.05` we have enough evidence to conclude that the true mean for the ages is significantly higher than 24

Explanation:

Information provided

`\bar X=25` represent the mean for the ages

`s=2` represent the sample standard deviation

`n=16` sample size

`\mu_o =24` represent the value to check

`\alpha=0.05` represent the significance level

t would represent the statistic

`p_v` represent the p value

Hypothesis to verify

We want to check if the average age of all the students at the university is significantly more than 24, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 24`

Alternative hypothesis:
`\mu > 24`

Since we don't know the population deviation the statistic would be given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the data given we got:

`t=(25-24)/((2)/(√(16)))=2`

The degrees of freedom are:

`df = n-1=16-1=15`

Now since we are conducting a right tailed test the p value can be calculated as follows:

`p_v =P(t_(15)>2)=0.0320`

Since the p value is lower than the significance level of
`\alpha=0.05` we have enough evidence to conclude that the true mean for the ages is significantly higher than 24