HOMEWORK

Homework

Question1

Thishypothesis test involves a sampling distribution of means that is astudent t distribution. The reason why the hypothesis test is astudent t distribution is because the number of observations beingtested is less than 30.

Question2

NullHypothesis H0: Brown = 0.13

AlternativeHypothesis H0: Brown ≠ 0.13

Question3

Z-statistic= (0.08 – 0.13) /sqrt (0.13) (1-0.13)/100

=-1.487

Question4

P-value= 0.1370

α= 0.05

Criticalvalue = 1.96 (two-tail test)

Thedecision concerning the null hypothesis is that the null hypothesisshould be accepted.

Question5

TheCritical value (1.96) is greater than the Test Statistic (-1.487)therefore, the null hypothesis should be accepted and the alternativehypothesis rejected. This is an indication that the claim that, ofthe Mars candy company the percentage of brown M&M’s is equalto 13%, is true.

Question6

NullHypothesis H0: µ = 30,000

AlternativeHypothesis HA: µ ≠ 30,000

Question7

Inthis case, the number of observations is less than 30 since thenumber of observations is 17 therefore, t-test statistic will beused.

Thetest statistic is equal to (22,298 – 30,000) / 14,200/sqrt 17

=-7702 / 3444.005

t= -2.24

Question8

Degreesof freedom = 17-1 = 16

Pvalue = 0.0399

Sincet is less than 2.120 there is sufficient evidence to reject theclaim that µ = 30,000. Thus, the decision of rejecting the nullhypothesis will be made.

Question9

Fromthe results, the value of t is less than 2.120, which is anindication that the null hypothesis should be rejected.

Question10

Degreesof freedom = 9-1 = 8

Criticalvalue = 3.355

Thecritical value is greater than the test statistic therefore, thenull hypothesis should not be rejected.

References

Wilcox,R. R. (2012). *Introductionto robust estimation and hypothesis testing*.London: Academic.