HWWeekbtTest Word MAILINGS REVIEW VIEW Foxit PDF E Emphasis Heading 1 1 NormalSt

Question

HWWeekbtTest Word MAILINGS REVIEW VIEW Foxit PDF E Emphasis Heading 1 1 NormalStrong Subtite T Title 1 No Spac. s Paragraph Styles 7. Complete a t-test using the data collected during the first week of class (Le, the question you asked classmates). You can compare groups via gender or major, depending on your hypothesis. For example, as I mentioned, last year an MLS student asked his classmates how many times they had seen Star Wars. He hypothesized that there was a significant difference between MLS and DEHS students, so he compared those two groups a. What type of inferential test should you use and why? Page 3 of 2 t test Homework Using the A-B-G-D format demonstrated in class, test the null hypothesis. (Use a s 05) b.

Explanation / Answer

ANSWER-

a) Here we use equality of two Sample t-test.

because we have given data is of the two groups

(how many pets have at male and female)

and we have to check wheather there is

significance difference in two groups or not.

b) we have given data based on how many

number of pets to have male and female.

Male 2,0,0,1,1,1,0,1,0

female 0,1,1,7,2,4,2,0,0,2,0,0,0,0,6,0,21,0,0,3,1

Here we have to test-

hypothesis for testing are -

H0: average number of pets at male and female is same.

against

H1: average number of pets at male and female is not same.

test statistic is

here population variance is unknown

thus we use sample variance

under H0,

by usinr R-software-

Rcode is-  

male=c(2,0,0,1,1,1,0,1,0)

female=c(0,1,1,7,2,4,2,0,0,2,0,0,0,0,6,0,21,0,0,3,1)

t.test(male,female)

OUTPUT-

>male=c(2,0,0,1,1,1,0,1,0)

> female=c(0,1,1,7,2,4,2,0,0,2,0,0,0,0,6,0,21,0,0,3,1)

> t.test(male,female)

Welch Two Sample t-test

data: male and female

t = -1.62, df = 21.993, p-value = 0.1195

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-3.9088413 0.4802698

sample estimates:

mean of x mean of y

0.6666667 2.3809524

HERE

p-value 0.1195 > 0.05

thus we accepte null hypothesis

and conclude that average number of pets at male and female is same.

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