According to one investigator’s model, the data are like 400 draws made at random from a large box. The null hypothesis says that the average of the box equals 50; the alternative says that the average of the box is more than 50. In fact, the data averaged out to 52.75, and the SD was 25. Compute z and P . What do you conclude?

If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the mean is significantly diffrent fro 50 at 1% of signficance.

Step-by-step explanation:

1) Data given and notation

[tex]\bar X=52.75[/tex] represent the mean height for the sample

[tex]s=25[/tex] represent the sample standard deviation

[tex]n=400[/tex] sample size

[tex]\mu_o =50[/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 check if the mean is equal to 50 or not, the system of hypothesis would be:

Null hypothesis:[tex]\mu = 50[/tex]

Alternative hypothesis:[tex]\mu \neq 50[/tex]

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value but we can assume it as z distribution, and the statistic is given by:

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

z-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]z=\frac{52.75-50}{\frac{25}{\sqrt{400}}}=2.2[/tex]

P-value

Since is a two sided test the p value would be:

[tex]p_v =2*P(Z>2.2)=0.0278[/tex]

Conclusion

If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the mean is significantly diffrent fro 50 at 1% of signficance.