Question

In: Statistics and Probability

Testing a Claim about a Mean Test the Claim: The mean time to take this exam...

Testing a Claim about a Mean Test the Claim: The mean time to take this exam is greater than 30 minutes. (Note: There will be NO credit given for wrong answers even if they are correct based on previous 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. (

Solutions

Expert Solution

We want to test that mean time to take this exam is greater than 30 minutes.

Here we are going to use one sample t test because population standard deviation is unknown.

This is one sided test.

We have given Population mean µ = 30 minutes

Sample mean (x) = 32 minutes

Sample standard deviation (s) = 5 minutes

Sample size (n) = 25

  1. Test statistics (t) = (x - µ) / (s/ √n)

t = (32 - 30)/ (5/sqrt(25))

Test statistics t = 2

  1. The one sided t critical value at degrees of freedom n-1 = 24 and α=0.05

T24, 0.05 = 1.711 (from statistical table)

Rejection region: Reject H0 if test statistics t > 1.711.

  1. Null hypothesis H0: µ = 30 min
  2. Alternative hypothesis H1: µ > 30 min

  1. Conclusion: t statistics = 2 > t24, 0.05 = 1.711 that means t statistics value fall in rejection region. So we reject null hypothesis at 5% level of significance.

We can conclude that, there is sufficient evidence of 0.05 level of significance that the mean time to take this exam is greater than 30 minutes.


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