Consider testing H^0: u=20 against H^a: u<20 where u is the mean number of latex gloves used per week by all hospital employees, based on the summary statistics n=46, x=19.1, and s=11.8. Complete parts a and b

a. Compute the p-value of the test.

The p-value of the test is _____ (Round to four decimal places as needed )

b. Compare the p-value with a=0.01 and make the appropriate conclusion. Choose the correct answer below.

a. There is insufficient evidence to reject h^0

b. There is sufficient evidence to reject H^0

If we compare the p value and the significance level given [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, and we can conclude that the true mean is not lower than 20 at 1% of signficance.

a. There is insufficient evidence to reject h^0

Step-by-step explanation:

Data given and notation

[tex]\bar X=19.1[/tex] represent the sample mean

[tex]s=11.8[/tex] represent the sample standard deviation

[tex]n=46[/tex] sample size

[tex]\mu_o =20[/tex] represent the value that we want to test

[tex]\alpha=0.01[/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 lower than 20, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 20[/tex]

Alternative hypothesis:[tex]\mu < 20[/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, 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{19.1-20}{\frac{11.8}{\sqrt{46}}}=-0.517[/tex]

Part a: P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n-1=46-1=45[/tex]

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

[tex]p_v =P(t_{(45)}<-0.517)=0.303[/tex]

Part b: Conclusion

If we compare the p value and the significance level given [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, and we can conclude that the true mean is not lower than 20 at 1% of signficance.

a. There is insufficient evidence to reject h^0