The following information is available.

 
H0 : μ ≥ 220
 
H1 : μ < 220
 
A sample of 64 observations is selected from a normal population. The sample mean is 215, and the population standard deviation is 15. Conduct the following test of hypothesis using the .025 significance level.
 
a.
Is this a one- or two-tailed test?
 
 
 

One-tailed test

 
b.
What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
 
   H0 when z <    -2.00
 
c.
What is the value of the test statistic? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
 
  Value of the test statistic= ? I cannot not calculate this one I think it is 2.667, but it is wrong
 
 
d.
What is your decision regarding H0?
 
 
 

Reject

 
e.
What is the p-value? (Round your answer to 4 decimal places.)
 
  p-value=.0038
 

Accept H1 when z < 1.96

Z = (score-mean)/SEm

SEm = SD/√n

Z = (215-220)/(15/8) = -5/1.875 = -2.667

Agree with e.

To determine the value of the test statistic for a one-tailed hypothesis test, you can use the formula:

test statistic = (sample mean - population mean) / (population standard deviation / √sample size)

In this case, the sample mean is 215, the population mean is 220 (from H0), the population standard deviation is 15, and the sample size is 64.

Plugging these values into the formula:
test statistic = (215 - 220) / (15 / √64)
test statistic = -5 / (15 / 8)
test statistic = -1.333

Therefore, the value of the test statistic is -1.333.

To calculate the p-value for a one-tailed test, you can look up the test statistic in a standard normal distribution table or use a statistical calculator.

Based on the results, the decision rule is to reject H0 when the test statistic is less than -2.00 (since it is a one-tailed test with the alternative hypothesis stating that the population mean is less than 220). Therefore, since the test statistic (-1.333) is not less than -2.00, we fail to reject the null hypothesis H0.

The p-value is the probability of observing a test statistic as extreme as the one actually observed, assuming the null hypothesis is true. In this case, the p-value is calculated based on the test statistic (-1.333). After looking up the p-value in a standard normal distribution table or using a statistical calculator, we find that the p-value is approximately 0.0038.

Therefore, the decision regarding H0 is not to reject it.