Question

Surveyed 30 people, 21 reported that they workout, 9 reported they do not

Studying if exercise helps to relieve stress and boast GPA in school.

Perform a hypothesis test on either a mean or proportion, and describe any potential errors of this test

Of the reported 22, mean GPA for students who workout is 3.62

standard deviation:

0.31016124838542

Null hypothesis: The population mean for GPA for those participants who work out is 3.62 (mean of GPA)

Alternative hypothesis: The population mean for GPA for those participants who work out does not equal 3.62, it could be higher or it could be lower

Significance level: 0.05

Perform a hypothesis test on a relationship, and describe any potential errors

Signficance of stress level in comparsion to amount of times person works out in a week

stress level given a number : low-1, moderate-2, high-3

person 1: 3 times a week, stress: 3

2: once a week, 2

3: 2x a week, 3

4: 4x a week, 3

5:6x a week, 2

6;once a week, 3

7:3x a week, 3

8:once a week, 2

9:once a week, 2

10:once a week, 2

11;4x a week, 2

12:once a week, 2

13:2x a week, 2

14:once a week, 2

15:3x a week, 1

16:6x a week, 1

17:2x a week, 2

18:2x a week, 3

19:once a week,2

20:3x a week, 3

21:2x a week, 3

Null hypothesis: Those who work out more often have more amounts of stress Alternative hypothesis: Those who work out have the same amount of stress significance level: 0.05

Answer 1

Solution:

1. Null Hypothesis (Ho): µ ? 3.1

Alternative Hypothesis (Ha): µ > 3.1

It is a right tailed test.

Test Statistics

Z = (X-bar - µ)/ (?/?n)

Z = (3.11 – 3.1)/ (0.05/?60)

Z = 1.55

Using Z-tables, the p-value is

P [Z > 1.55] = 0.06

Since p-value is greater than 0.025 level of significance, we fail to reject Ho.

2. Null Hypothesis (Ho):µ = 3.5

Alternative Hypothesis (Ha): µ ? 3.5

Test Statistics

Z = (X-bar - µ)/ (?/?n)

Z = (3.52 – 3.5)/ (0.08/?50)

Z = 1.77

Using Z-tables, the p-value is

P [Z ? 1.77] = 0.0767

Since p-value is less than 0.20 significance level, we reject Ho.

3. Test Statistic

Z = (X-bar - µ)/ (?/?n)

Z = (7.8- 9 )/ (4.8/?80)

Z = -2.236

Using Z-tables, the p-value is

P [Z < -2.236] = 0.0127

Since p-value is less than 0.10 significance level, we reject Ho.

The sample data support the claim that the population mean is less than 9.

Solution 4:

Test Statistic

t = (x-bar - µ)/ (s/?n)

t = (81 – 70.7)/ (11.3/?7)

t = 2.412

Degrees of freedom, df = n – 1= 7 – 1 = 6

Using t-tables, the p-value is

P [t (6) > 2.412] = 0.0262

Since p-value is greater than 0.001 significance level, we fail to reject Ho.

There is sufficient evidence to warrant rejection of the claim that the population mean is greater than 70.7.