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The following data was collected from a group of students who are taking a statistics course: Subject # gender breakfast Sleep Hrs Study Hrs Quiz Faculty 1 f n 7 6 8 arts 2 f y 7 5 12 med 3 f n 9 5 9...

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The following data was collected from a group of students who are taking a statistics course:

Subject #

gender

breakfast

Sleep Hrs

Study Hrs

Quiz

Faculty

1

f

n

7

6

8

arts

2

f

y

7

5

12

med

3

f

n

9

5

9

arts

4

f

n

6.5

8

7

arts

5

f

n

8

8

12

arts

6

m

y

8

8

8

arts

7

f

y

6.5

6

10

sci

8

f

n

6.5

6

3

sci

9

f

n

7

4.5

3

arts

10

f

y

7.5

8.5

9

sci

11

m

n

7

4

8

arts

12

m

n

9

5

9

arts

13

m

n

6.5

8.5

8

arts

14

m

y

7

2

9

sci

Specify which test would be used for each of the following hypotheses.

  1. 1) Males and females differ in whether or not they regularly eat breakfast.

  2. 2) The breakdown of students into faculty in PSYC 2020 is not uniform.

  3. 3) Students invest more hours in sleep each night than they do in stats all week.

  4. 4) Study hours depend on quiz scores.

  5. 5) Students in Arts and Science perform differently on the quiz.

Conduct the tests for #3 and #5 and write a short report for each.
For #1, #2 and #4, state only H0and H1and df and critical value, but no calculation is necessary.

Answered Same DayNov 24, 2021

Solution

Pooja answered on Nov 25 2021
50 Votes
1)
Null hypothesis, Ho: gender and
eakfast are independent of each other.
Alternative hypothesis, H1: gender and
eakfast are dependent on each other.
Df = (r-1)*(c-1) = (2-1)*(2-1) = 1
Critical value, chi-square (0.05,1) =CHISQ.INV.RT(0.05,1) = 3.841
2)
Null hypothesis, ho: there are equal proportion of students in each category of faculty namely arts, medical and science. P1 = p2 = p3 = 1/3 = 0.33
Alternative hypothesis, h1: at least one of the proportion of students in each category of faculty namely arts, medical and science is different. At least one p1 =/= p2 =/= p3 =/= 0.33
K=3 (number of categories)
Df = (k-1) = 3-1 = 2
Critical value, chi-square (0.05,2) =CHISQ.INV.RT(0.05,2) = 5.991
3)
Null hypothesis, Ho: there is no significant difference in the mean sleep hours and stat study hours. U1=u2
Alternative hypothesis, h1: the mean sleep hours is more than mean stat study hours. U1>u2
Alpha = 5%
The groups of sleep hours and study hours are related...
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