Hypothesis Testing problems (please show work) For ...
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 p-value? 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
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