Question

John was studying how far the UMN students travelled to take the class on campus. Let W be the distance travelled (measured in miles). A sample of size 25 was collected. He calculated the following statistics: John wanted to run a hypothesis test with the null and alternative hypothesis as below: Ho : μ = 5 Note that μ is the mean distance traveled by UMN students to come to campus. Please calculate the test statistic for the hypothesis test. Please round the answer to the nearest hundredth.

Answer 1

The given information is as shown below: Sample size, n= 25 W = 3.8 25 E(w, -w) = 23 i=1 The null and alternative hypotheses can be defined as, H: µ= 5 Vs H:u #5 The test statistic is, W-4 s/n Here, W be the sample mean, uis the population mean, s is the sample standard deviation and nis the sample size. Thus, 25 (w,-w n-1 i=1 (23) 25 –1 = 0.9789 %3D

Thus, 3.8 – 5 0.9789/25 -1.2 0.19578 =-6.12933 x -6.129 is -6.129 Therefore, the t-sta The degrees of freedom is, df =n-1 = 25 –1 = 24 The alternative hypothesis is two tail, so the P-value using excel function is as shown below: P-value = TDIST (|t|. df ,tail) = TDIST (6.129,24,2) = 0.0000 Here, the P-value is very low; so reject the null hypothesis.