4.) Using the level of significance of .05, test the claim that the average numb

Question

4.) Using the level of significance of .05, test the claim that the average number of classes skipped by CLC students is less than 5. The sample data used to test the claim is as follows: 2, 3, 6, 4, 1, 1, 7, 4, 3, 2, 0, 5, 0, 1, 2, 4, 0, 3, 6, 1 a. Find the mean and sample standard deviation using Excel Homework 1 information b. Find the test value using TTest on your calculator. c. Find the p-value using T.dist on Excel d. Find the critical value using T.inv on Excel e. Type all 7 hypothesis testing steps

Explanation / Answer

Here we want to test the hypoyhesis that the average number of classes skipped by clc student is less than 5.

i.e. Ho: mu=5 against H1: mu<5 ( left-tailed test)

Since sample size =20(<30) and population variance is unknown we use one sample t-test

under Ho the test statistic is

t = ( xbar-mu)/( s/sqrt(n) ~ tn-1

t =( 2.75-5)/( 2.1244194/sqrt(20))

=-4.7364

to find p-value with t=4.7364 , d.f=19 and size is one tailed

p-value= 0.0000728

Critical value : alpha = level of significance =0.05 and d.f =19

ttab= 2.0930

Decision criteria :

for left tailed test - t tab > tcal then we reject Ho at alpha % level of significance.

here -2.0930 > -4.7364

Hence we reject Ho at 5% level of significance.

Conclusion : The average number of classes skipped by clc students is less than 5.

Sample no. Number of classes skipped by students 1 2 Mean=xbar average(B2:B21) 2.75 2 3 S.D=s stdev(B2:B21) 2.1244194 3 6 t (xbar-mu)/(s/sqrt(n) -4.736497 4 4 p-value tdist(4.73,19,1) 7.286E-05 5 1 critical value tinv(alpha,d.f) 2.093024 6 1 7 7 8 4 9 3 10 2 11 0 12 5 13 0 14 1 15 2 16 4 17 0 18 3 19 6 20 1

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