Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use the P-Vakue metho?
Assume that a simple random sample has been selected from a normally distributed population and test the given claim.
Use the P-Vakue method.
Below, dentify:
a) The null hypothesis
b) The alternative hypothesis
c) The test statistic
d) The P-value
e) The critical value(s)
f) The conclusion about the null hypothesis
g) The final conclusion that addresses the original claim.
A researcher wants to test the claim that convicted burglars spend an average of 18.7 months in jail. She takes a
random sample of 11 such cases from court files and finds that x = 20.6 months and s = 7.8 months. Test the
claim that µ = 18.7 months at the 0.05 significance level. Use the traditional method of testing hypotheses.
1 Answer
- paramvenuLv 76 years ago
a) Ho: Mu = 18.7
b) Ha: Mu is NOT = 18.7 (two tailed test)
[if it is Mu > 18.7 then it is a right tailed test]
c) Test statistic = t = [Xbar - Mu] / [s/sqrt n]
= [20.6 - 18.7] / [7.8/sqrt 11]
= + 0.808
d) Online p-value calculator for t-test is used which is available FREE on making a Google search
Degrees of freedom = n-1 = 11-1 = 10
p-value = 0.4379 for two tailed test
0.2189 for right tailed test
e) Critical value of t for 10 df at alpha = 0.05 is 2.228 for two tailed test and
1.812 for right tailed test
f) Since p-value (0.4379) > the alpha value (0.05) Ho will be accepted
Since the test statistic t-value (0.808) < the critical value of t (2.228) Ho will be accepted
g) It is inferred that the test supports the claim that Mu = 18.7
