Question

A recent study showed that the average number of sticks of gum a person chews in a week is 11. A college student believes that the guys in his dormitory chew less gum in a week. He conducts a study and samples 24 of the guys in his dorm and finds that on average they chew 17 sticks of gum in a week with a standard deviation of 3.8. Test the college student's claim at α =0.10. The correct hypotheses would be: H 0 : μ ≤ 11 H A : μ > 11 (claim) H 0 : μ ≥ 11 H A : μ < 11 (claim) H 0 : μ = 11 H A : μ ≠ 11 (claim) Since the level of significance is 0.10 the critical value is -1.32 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: reject H 0 do not reject H 0 The final conclusion is that: There is enough evidence to reject the claim that less gum in a week. There is not enough evidence to reject the claim that less gum in a week. There is enough evidence to support the claim that less gum in a week. There is not enough evidence to support the claim that less gum in a week.

Answer 1

Sol:

Ho:mu=11

Ha:mu<11

H 0 : μ ≥ 11 H A : μ < 11

df=n-1=24-1=23

t crirical value in excel

=T.INV(0.1,23)

=-1.31946024

=-.1.32

Test statistic

t=xbar-mu/s/sqrt(n)

=(17-11)/(3.8/sqrt(24))

= 7.735231

=7.735

test statistic,t=7.735

pdf=n-1=24-1=23

p value in excel is

=T.DIST( 7.735231,23,TRUE)

=1

Do not reject Ho

There is not enough evidence to support the claim that less gum in a week.

H 0 : μ ≥ 11 H A : μ < 11

t=-.1.32

test statistic,t=7.735

p value=1.000

do not reject H 0

There is not enough evidence to support the claim that less gum in a week.