Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

metropo lan spotaton u on N has se bus mec anal reia0 ly 225 basis comple parts

ID: 3364431 • Letter: M

Question

metropo lan spotaton u on N has se bus mec anal reia0 ly 225 basis comple parts (A) andana aal o 900 bus mi es us mechanica rei is measured s ecfica v os ne nur be on bus mi es et een mechanica oad cals Suppose 35amo e of 00 buses resubed sa ple mean o 950 ou5 miles nd sa p e stand rd deviation o Sate the null and atemative hypotheses Type or decimal ha critikcal vauat8) fartha tast statistic iaan there fficiant evidence to r ectthe nuil hypohcsls uing -0.05? O A. Do not reec: t e nul hypothes s. Ther", "su rat t ev ence yt the 0.05 evt sig cance th . the po ultr on mean bus mies 15 greater than 3.500 bus mies O B. Do not rejec: thc nul hypothesis. Tharensufficient cvidence t the 0.05 levzl of sig.ificance that the populaton mean bus mies i 125 t n 3.90 bus mies O C. Raject te nul nypotnesls. I hare s sutticlontevidarce atte U0b level oesgnncancatat the population mean bus mlas less than 3 yuu bus mies. What does this p valus mcan given the rezultz af part a? O A. 'ha 0-valua is tha prohatilty ct gat! ng a sampia maan at 3950 hus mías ar graatar it the actual mean 390d bus mias OD. The pvalue is the probability that the sctual msan i: greate than 3,950 bus miles.

Explanation / Answer

x = 3950

= 225

Ho : µ 3900 / µ = 3900

H1 :µ > 3900

The test statistic is:

t = (x - µ) / (s/n) = (3950-3900)/ (225/10) = 2.222

df= n-1 = 100-1 = 99

Critical value = t0.05,99 = 1.66

Since critical value (1.66) < t(2.222), reject the null hypothesis.There is sufficient evidence at the 0.05 level of significance that the population mean bus miles is greater than 3900 bus miles.

Using R software,

> pt(-2.222,df=99) #p-value

[1] 0.01428

The p-value is the probability that the actual mean is 3950 bus miles or less.