Question

Suppose a new construction company claims that, the mean time to complete a commercial complex is around 15 months. A sample of 41 construction companies are randomly selected and it is found that the mean time taken by them to complete a commercial complex building was 12.5 with a standard deviation of 2.5 months. At 1% level of significance, do you have sufficient evidence to conclude that the new construction company claim true?

What is the Null hypothesis (H0); Alternate hypothesis (H1); and the direction of the test?                            [ Select ]                       ["mu=15; mu<15; Left tail Test", "mu=15; mu<>15; Two tail Test", "mu=15; mu>15; Right tail Test", "x-bar=15; x-bar<>15; Two tail Test"]      

Which distribution is used in this case?                            [ Select ]                       ["z", "Both", "t with 40 d.f.", "Not Sure"]      

What is the critical value?                            [ Select ]                       ["2.58", "2.7045", "2.7045 and -2.7045", "2.58 and -2.58"]      

What is the Test Statistic Value?                            [ Select ]                       ["-6.4", "6.32", "-6.32", "6.4"]      

What is the P-value?                            [ Select ]                       ["pvalue<0.0001", "0.001", "1", "0.0002"]      

What is the conclusion of the test?

Answer 1

The answers are:

• "mu=15; mu<>15; Two tail Test
• t with 40 d.f
• "2.7045 and -2.7045"
• "-6.4"
• "pvalue<0.0001"
• Reject H0

Calculations:

One-Sample t-test

The sample mean is Xˉ=12.5, the sample standard deviation is s=2.5, and the sample size is n=41. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ =15 Ha: μ ≠15 This corresponds to a Two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2a) Critical Value Based on the information provided, the significance level is α=0.01, and the degree of freedom is n-1=41-1=40. Therefore the critical value for this Two-tailed test is tc​=2.7045. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Two-tailed test is |t|>2.7045 i.e. t>2.7045 or t<-2.7045 (3)Test Statistics The t-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |t|=6.4031 > tc​=2.7045, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0, and since p=0≤0.01, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 15, at the 0.01 significance level.