Q1. Suppose that in a one-tail hypothesis test where you reject H0 only in the upper tail, you compute the value of the test statistic Z to be 2.00. What is the p value?
pnorm(2.00,0,1,lower.tail=FALSE)
Answer: 0.02275013
Q2. Suppose that in a one-tail hypothesis test where you reject H0 only in the lower tail, you compute the value of the test statistic Z to be –1.38. What is the p value?
pnorm(-1.38,0,1,lower.tail=TRUE)
Answer: 0.08379332
Q3. Suppose that in a one-tail hypothesis test where you reject H0 only in the lower tail, you compute the value of the test statistic Z to be +1.38. What is the p value?
pnorm(1.38,0,1,lower.tail=TRUE)
Answer: 0.9162067
Q4. Dataset: Hourly_Wage. The data accompanying this exercise show hourly wages (Wage in $) for 50 employees. An economist wants to test if the average hourly wage is less than $22. Assume that the population standard deviation is $6.
summary(Hourly_Wage) nrow(Hourly_Wage)
4a. State the null and the alternative hypotheses for the test.
\(H_0:\mu \ge 22\); \(H_1:\mu<22\)
4b. Find the value of the test statistic and the p-value.
SE=6/sqrt(50)
ZT= (23.15-23)/SE ZT= 0.176776695296635
pvalue<- pnorm(ZT,0,1,lower.tail=TRUE)
P-Value= 0.5701581
4c. At 0.05 significance level, what is the conclusion to the test? Is the average hourly wage less than $22?
Since p_value (0.5701581) > \(\alpha\) (0.05), do not reject the null hypothesis. At the 5% significance level, we have insufficient evidence to conclude average hourly wage is less than $22. The sample data does not support the alternative hypothesis.