Question

a consultant for a large university studied the number of hours per week freshmen watch tv versus the number of hours seniors do. the results of this study follow. is there enough evidence to show the mean number of hours per week freshmen watch tv is different from the mean number of hours seniors do at alpha=.1. freshmen n=10, xbar=18.6, s=7.8740 seniors n=4, xbar=11.4, s=3.9749

Answer 1

SOLUTION:

From given data,

	Freshmen	Seniors
Mean ()	= 18.6	= 11.4
SD (s)	= 7.8740	=3.9749
var (s2)	= 61.999876	= 15.79983001
n	= 10	= 4

Null hypothesis , H₀ :

Alternative hypothesis , H₁ :

Level of significance , = 0.10

Finding the degree of freedom .

df = ( / + /) / [(( / )² / -1) +(( /)² / -1)]

= (61.999876 / 10 + 15.79983001 /4) / [((61.999876/10)² / 10-1) +((15.79983001 /4)² / 4-1)]

= 10.1499451025 / (4.2710940 + 5.2007214238)

= 10.1499451025 / 9.4718154238

= 1.07159

= 2

Find the critical value ,

t_critical = (= TINV( , df)) (Use MS Excel function)

   (TINV(0.10 , 2))

= + 2.91998

Decision : Reject H₀ ; when t_critical > 2.91998 (OR)

Decision : Reject H₀ ; when p < 0.10

The Test statistics is,

t = 2.2599

Find the p- value

P- value = (= TDIST(2.2599 , 2)) (Use MS Excel function)

= 0.15230

Conclusion:

Critical value approach:

The test statistic value is less than the critical value , so we fail to reject the null hypothesis.

p-value approach:

The p - value is greater than the significance level 0.10, so we fail to reject the null hypothesis,

Hence , the p - value is not lies between , 0.05 < P < 0.10