2. A particular resource class must perform at a level consistent with the rest of the school to avoid being discontinued. On the first quarter statewide test, the mean score of the school was 78. The 14 students in the resource class had a 71.86 mean test score and a 17.52 standard deviation. The district informs the principal that the class should be discontinued because the average is below that of the rest of school. Is this reasonable or does the principal have evidence to support an argument against this decision

If we compare the p value and a significance level assumed for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the scores is NOT significant less than 78.

Step-by-step explanation:

Data given and notation

[tex]\bar X=71.86[/tex] represent the sample mean

[tex]s=17.52[/tex] represent the standard deviation for the sample

[tex]n=14[/tex] sample size

[tex]\mu_o =78[/tex] represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is less than 78, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 78[/tex]

Alternative hypothesis:[tex]\mu < 78[/tex]

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{71.86-78}{\frac{17.52}{\sqrt{14}}}=-1.311[/tex]

Calculate the P-value

First we need to find the degrees of freedom given by:

[tex]df=n-1=14-1=13[/tex]

Since is a one-side left tailed test the p value would be:

[tex]p_v =P(t_{13}<-1.311)=0.106[/tex]

Conclusion

If we compare the p value and a significance level assumed for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the scores is NOT significant less than 78.