Question

A professor wants to know if her introductory statistics class has a good grasp of basic math. Six students are chosen at random from the class and given a math proficiency test. The professor wants the class to be able to score above 70 on the test. The six students get scores of 62, 92, 75, 68, 83, and 95. Can the professor state that the mean score for the class on the test would be above 70?

Answer 1

`df=n-1=6-1=5`

`p_v =P(t_((5))>1.706)=0.0694`

If we select a confidence level of
`\alpha=0.05` (commonly used significance level) we see that
`p_v>\alpha` so then we FAIL to reject the null hypothesis and we can't conclude that the true mean is higher than 70. If we select any significance level lower than 0.06 we reject the null hypothesis and we will have enough evidence to reject the null hypothesis.

Explanation:

Data provided

62, 92, 75, 68, 83, and 95.

We can calculate the sample mean and deviation with the following formulas:

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

`s=\sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

`\bar X=79.17` represent the sample mean for the student scores

`s=13.17` represent the sample standard deviation

`n=6` sample size

`\mu_o =70` represent the value that we want to test

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean for the scores is above 70, and the hypothesis for this case are:

Null hypothesis:
`\mu \leq 70`

Alternative hypothesis:
`\mu > 70`

Since we don't know the population deviation the statstic for this case is:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing in formula (1) we got:

`t=(79.17-70)/((13.17)/(√(6)))=1.706`

Conclusion

We need to find first the degrees of freedom like this:

`df=n-1=6-1=5`

And the p value would be:

`p_v =P(t_((5))>1.706)=0.0694`

If we select a confidence level of
`\alpha=0.05` (commonly used significance level) we see that
`p_v>\alpha` so then we FAIL to reject the null hypothesis and we can't conclude that the true mean is higher than 70. If we select any significance level lower than 0.06 we reject the null hypothesis and we will have enough evidence to reject the null hypothesis.