Question

Question (4) [6 marks]

I am a firm believer in sketching things on tests/assignments. I think that sometimes the sketch can help remind you what you have to do next, or correctly guide your thought process. One of my TA's for a previous Statistics course gathered data from marked tests to determine if sketching had an influence grades. For each student in the class, we recorded whether or not they sketched during a specific question, and also recorded whether or not they obtained the correct p- value. The results were as follows:

At 10% the level of significant, is there any statistical reason to believe that sketching is associated with grades?

Note: You can use the functions qchisq() in R to help you in solving the following. Why we are using qchisq() function in R.

The qchisq() function in R allows us to specify a desired area in a tail and the number of degrees of freedom. From that information, qchisq() computes the required x-value to get the specified area in the specified tail with the specified number of degrees of freedom.

O	sketch	No sketch	Total
correct	50	30
incorrect	60	90
Total

E	sketch	No sketch	Total
correct
incorrect
Total

1. a) State the two hypothesis of interest.

2. b) Calculate an appropriate test statistic for (a) by hand.

χ 2   (O  E )2 E

c) Write your conclusion using the rejection region method “critical value method” include both statistical and related to the topic of the question (practical) interpretation use the function qchisq() in R.

Question 5

In this question we’ll use housetasks dataset from STHDA: http://www.sthda.com/sthda/RDoc/data/housetasks.txt. The dataset is a contingency

table containing 13 house tasks and their distribution in the couple: → rows are the different tasks

→ values are the frequencies of the tasks done :

1) by the wife only
2) alternatively
3) by the husband only 4) or jointly

Using R test whether the two variables housetasks and their distribution in the couple are statistically significantly associated (dependent) by answering the following questions.

a)

b)

(1 mark) State the two hypothesis of interest.

(0.5 mark) Import the data into R using the function read.table

Note: show your R codes but not the output (the dataset).

c)

(0.5 mark) Calculate Chi-square statistic using the function chisq.test() in R. Note: show your R codes and output.

d)

(1 mark) Use ? = 0.05 write your conclusion using the p-value (include both statistical and related to the topic of the question interpretation).

DATA for Q5

            Wife        Alternating     Husband Jointly
Laundry   156   14      2       4
Main_meal 124   20      5       4
Dinner    77    11      7       13
Breakfeast      82      36      15      7
Tidying    53   11      1       57
Dishes    32    24      4       53
Shopping        33      23      9       55
Official        12      46      23      15
Driving 10      51      75      3
Finances        13      13      21      66
Insurance       8       1       53      77
Repairs 0       3       160     2
Holidays        0       1       6       153

Answer 1

a) Null hypothesis, H₀: sketching and grades are independent

vs. Alternative hypothesis, H_a: sketching has an influence on grades

(b)

O	sketch	No sketch	Total
correct	50	30	80
incorrect	60	90	150
Total	110	120	230

E	sketch	No sketch	Total
correct	80*110/230=38.2609	80*120/230=41.7391
incorrect	150*110/230=71.7391	120*150/230=78.2609
Total

Q5: (a) Null hypothesis, H₀: Two variables house tasks and their distribution in the couple are not associated

vs. Alternative hypothesis, H_a: Two variables house tasks and their distribution in the couple are associated

(b) and (c) R code:

library(MASS)       # load the MASS package
t=matrix(c(156,124,77,82,53,32,33,12,10,13,8,3,0,14,20,11,36,11,24,23,46,51,13,1,3,1,
2,5,7,15,1,4,9,23,75,21,53,160,6,4,4,13,7,57,53,55,15,3,66,77,2,153),ncol=4,byrow=TRUE)
colnames(t)=c("Wife","Alternating","Husband","Jointly")
rownames(t)=c("Laundry",
"Mainmeal",
"Dinner",
"Breakfeast",
"Tidying",
"Dishes",
"Shopping",
"Official",
"Driving",
"Finances",
"Insurance",
"Repairs",
"Holidays"
)
table=as.table(t)
chisq.test(table)

Output:

Pearson's Chi-squared test

data: table
X-squared = 809.81, df = 36, p-value < 2.2e-16

Warning message:
In chisq.test(table) : Chi-squared approximation may be incorrect

(d) Since p-value<0.05 so we reject H₀ at 5% level of significance and conclude that the two variables housetasks and their distribution in the couple are statistically significantly associated.