A large university is interested in learning about the average time it takes students to drive to campus. The university sampled 51 students and asked each to provide the amount of time they spent traveling to campus. The sample results found that the sample mean was 23.243 minutes and the sample standard deviation was 20.40 minutes. It is desired to determine if the population standard deviation exceeds 20 minutes. Calculate the test statistic for this test of hypothesis.A) x2 = 53.06 B) x2 = 58.11C) x2 = 52.02 D) x2 = 51

Answer: C) [tex]\chi^2=52.02[/tex]Step-by-step explanation:Given : It is desired to determine if the population standard deviation exceeds 20 minutes. i.e. [tex]H_a: \sigma>20[/tex]Here, we perform chi-square test.Sample size : n= 51Sample standard deviation : s= 20.40The test statistic for chi-square test is given by :-[tex]\chi^2=\dfrac{(n-1)s^2}{\sigma^2}[/tex], where n= sample size s= sample standard deviation.[tex]\sigma[/tex]= Population standard deviation.Substitute the values , we get[tex]\chi^2=\dfrac{(51-1)(20.40)^2}{(20)^2}\\\\=\dfrac{50\times 416.16}{400}=52.02[/tex]Hence, the test statistic for this test of hypothesis= [tex]\chi^2=52.02[/tex]Thus , the correct answer is C) [tex]\chi^2=52.02[/tex]