Statistics Exam Questions

Questions on hypothesis testing

Question 1- Match

a- 1

b-4

c-2

d-3

Question 2

Option E (None of the Above)

Question 3

Expected count= (16+15+10)*(10+1+5)/(16+15+10+10+6+1+4+2+5) = 9.5

Question 4

Option B

Question 5

Option A

Question 6

Option A

Question 7

Option C

Question 8

Option A

Question 9

Option E

Question 10

Option E

Question 10

Yes, Bobo’s claim is supported since the 95% confidence interval lies between 1.62 and 3,21 minutes.

Question 11

Given weight of 31, 29, 26, 33, 40, 28, 30, and 25, at α = 0.05 test the population mean is 35 – Option 2  (2-tailed test of mean with both negative and positive critical value)

Data were gathered in an experiment comparing the effect of three insecticides in controlling a certain species of parasitic beetle. Each observation represents the number of such insects found dead in a certain fixed area treated with an insecticide – Option 6 (ANOVA)

A multinomial probability distribution describes the distribution of counts across multiple level of a variable. A special case is a binomial discrete probability distribution. For each level of a variable, which is common to multiple populations, equality of distributions can be tested. – Option 7(Chi Square test of homogeneity)

Run times (msec) of a new phone app developed by GotYourGP and established competitor WeTrackYou.Inc. are logged in a data file. At α = 0.05, are the run time of GotYourGp and WeTrackYou.Inc. comparable? – Option 5 (2-tailed test of means of 2 independent populations)

Testing that a gasoline additive increases millage, we compute mpg with additive for each car. If the mean of the difference is positive enough, action will be required to implement the additive. – Option 4 (1-tailed test of paired-data with positive critical value)

Does the data validate the normal probability distribution assumption? –Option 9 (Chi-square test of goodness of fit)

Test size C battery mean life is at least 25 hours –Option 3 (1-tailed test of mean with negative critical value)

Do contingency table classification values matter?- Option 8(Chi-square test of independence)

Question 12

Total observations = 48

Observations where air quality has been met = 40

Proportion meeting air standards = (40/48) = 0.8333

Hypothesised proportion = 0.9

Null Hypothesis: π ≥ 0.90

Alternative Hypothesis: π < 0.90

Z statistic = (0.8333-0.9)/(0.9*0.1/48)^0.5 = -1.5396

The corresponding p value for the above z statistic = 0.0618

Since the p value is less than significance level of 10%, hence the null hypothesis is rejected and alternative hypothesis is accepted. Hence, it can be concluded that air quality meets standards less than 90% times.

Question 13

Option E

Question 14

Option D

Question 15

Option B