Hypothesis Testing problems (please show work) For Exercises 1–4 answer the
questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is
the value of the test statistic? (d) What is your decision regarding H0? (e) What is the pvalue? Interpret it.
1. The following information is available. H0: µ= 50 H1: µ≠ 50 The sample mean is 49,
and the sample size is 36. The population standard deviation is 5. Use the .05 significance
level.
a) This is a two-tailed test.
b) Reject H0 if the absolute value of the test statistic |z| > 1.96
c) The test statistic, z = (xbar – 50)/(σ/√n) = (49-50)/(5/√36) = -1/(5/6) = -1.2
d) Here, |z| = 1.2 < 1.96. So we fail to reject H0.
e) The p-value = P[|Z|> 1.2] = 0.2301
2. The following information is available. H0: µ= 10 H1: µ> 10 The sample mean is 12
for a sample of 36. The population standard deviation is 3. Use the .02 significance level.
a) This is a one-tailed test.
b) Reject H0 if the value of the test statistic z > 2.054
c) The test statistic, z = (xbar – 10)/(σ/√n) = (12-10)/(3/√36) = 2/(3/6) = 4
d) Here, z = 4 > 2.054. So we reject H0.
e) The p-value = P[Z > 4] = 0.0000
3. A sample of 36 observations is selected from a normal population. The sample mean is
21, and the sample standard deviation is 5. Conduct the following test of hypothesis using
the .05 significance level. H0: µ= 20 H1: µ> 20
a) This is a one-tailed test.
b) Reject H0 if the value of the test statistic t > 1.690
(From Student’s t distribution with n-1 = 35 degrees of freedom]
c) The test statistic, t = (xbar – 20)/(s/√n) = (21-20)/(5/√36) = 1/(5/6) = 1.2
d) Here, t = 1.2 < 1.690. So we fail to reject H0.
e) The p-value = P[t > 1.2] = 0.1191
4. A sample of 64 observations is selected from a normal population. The sample mean is
215, and the sample standard deviation is 15. Conduct the following test of hypothesis
using the .03 significance level. H0: μ≥220 H1: μ< 220
a) This is a one-tailed test.
b) Reject H0 if the value of the test statistic t < -1.915
(From Student’s t distribution with n-1 = 63 degrees of freedom]
c) The test statistic, z = (xbar – 220)/(s/√n) = (215-220)/(15/√64) = -5/(15/8) = -2.6667
d) Here, t = -2.6667 < -1.915. So we reject H0.
e) The p-value = P[t < -1.915] = 0.0049