Testing a Claim about a Mean T
Testing a Claim about a Mean Test the Claim: The mean time totake this exam is greater than 30 minutes. (Note: There will be NOcredit given for wrong answers even if they are correct based onprevious wrong answers – so check your work) Sample data: n = 25, x= 32 minutes, s = 5 minutes. α = 0.05
6. What is the value of the Test Statistic? (1 Point)
7. What is/are the Critical Value(s)? (1 Point) Ho:
8. Null Hypothesis. (1 Point) H1:
9. Alternate Hypothesis. (1 Point)
10. Reject or Fail to Reject. (
Answer:
We want to test that mean time to takethis exam is greater than 30 minutes.
Here we are going to use one sample ttest because population standard deviation is unknown.
This is one sided test.
We have given Population mean µ = 30minutes
Sample mean (x) = 32 minutes
Sample standard deviation (s) = 5minutes
Sample size (n) = 25
- Test statistics (t) = (x – µ) / (s/ √n)
t = (32 – 30)/ (5/sqrt(25))
Test statistics t = 2
- The one sided t critical value at degrees of freedom n-1 = 24and α=0.05
T24, 0.05 = 1.711 (fromstatistical table)
Rejection region: Reject H0if test statistics t > 1.711.
- Null hypothesis H0: µ = 30 min
- Alternative hypothesis H1: µ > 30 min
- Conclusion: t statistics = 2 > t24, 0.05 = 1.711that means t statistics value fall in rejection region. So wereject null hypothesis at 5% level of significance.
We can conclude that, there issufficient evidence of 0.05 level of significance that the meantime to take this exam is greater than 30 minutes.
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