Question

As a class research project, Rosaura wants to see whether the stress of final exams elevates the blood pressures of freshmen women. When they are not under any untoward duress, healthy eighteen-year-old women have systolic blood pressures that average 120 mm Hg with a standard deviation of 12 mm Hg. If Rosaura finds that the average blood pressure for the fifty women in Statistics 101 on the day of the final exam is 125.2, what should she conclude?

Answer 1

`z=(125.2-120)/((12)/(√(50)))=3.06`

`p_v =P(Z>3.06)=0.0011`

If we compare the p value and the significance level assumed
`\alpha=0.01` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean for the blood pressure in the class is significantly higher than 120 mm Hg at 1% of signficance.

Explanation:

1) Data given and notation

`\bar X=125.` represent the mean for the blood pressure sample

`\sigma=12` represent the population standard deviation assumed

`n=50` sample size

`\mu_o =120` represent the value that we want to test

`\alpha` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` 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 height blood pressure is higher than the average of 120 mm Hg, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 120`

Alternative hypothesis:
`\mu > 120`

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (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:

`z=(125.2-120)/((12)/(√(50)))=3.06`

P-value

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

`p_v =P(Z>3.06)=0.0011`

Conclusion

If we compare the p value and the significance level assumed
`\alpha=0.01` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean for the blood pressure in the class is significantly higher than 120 mm Hg at 1% of signficance.