For this project, use the class data to complete some hypothesis tests. Assume t
ID: 3067344 • Letter: F
Question
For this project, use the class data to complete some hypothesis tests. Assume this data is representative of all online students. For each hypothesis test, report the following:
The null hypothesis, H0
The alternative hypothesis, H1
The test statistic rounded to the nearest hundredth
The P value for the test
The formal decision (Reject H0 or Fail to reject H0)
The conclusion of the test in non-technical terms
Remember that each of these test is either a test for the population proportion, p, or a test for the population mean, mu. You have to decide which is which and set up the hypotheses accordingly. Once you know what parameter you are testing (p or mu), then that determines which function you should use in StatCrunch to find the test statistic (either a z or a t value) and the P value for the test. Each problem is worth a total of 6 points (1 point for each correct test component). You can also earn one point for rounding as directed.
1) Use a 5% significance level to test the claim the average online student got less than 8 hours of sleep last night.
2) Use a 1% significance level to test the claim a majority of online students will call "heads" will given the option to call a coin toss.
3) Use a 5% significance level to test the claim the mean foot length for online students is less than 25 cm.
4) Use a 5% significance level to test the claim the mean number of letters in the last names of online students is different from 6 letters (the most common last name size according to https://www.quora.com/What-is-the-average-length-of-last-names-in-the-United-States (Links to an external site.)Links to an external site.).
5) Use a 5% significance level to test the claim that less than 25% of online students think cats make the best pet.
6) Use a 5% significance level to test the claim the proportion of online students that pick the number 5 when asked to pick a number between 1 and 5 is different from 20%.
7) Use a 1% significance level to test the claim the proportion of online students with blue eyes is 8%. Note, the website Error! Hyperlink reference not valid.reports about 8% of the world's population has blue eyes.
Hours of Sleep
Coin Toss Call
T
H
T
H
T
H
T
H
H
H
H
H
T
T
H
H
H
T
H
H
T
H
T
T
H
T
H
H
H
H
T
H
H
T
T
H
T
T
T
H
H
T
H
H
T
H
T
H
H
H
H
H
H
T
H
3.75
8
6.25
7
8
8.75
8
5
6.5
7.25
9
8.75
5
8.25
8
5.5
6
8.5
7
6.5
6.5
8.25
5.75
8
6.25
6
5.25
8
7
9
6.5
9.5
5.5
6
8
6
8
8
8.5
6
4
6
8.25
8
10.5
7
8.75
5.5
5.5
6
2.5
9
8
6.5
6.75
Foot Length (cm)
26
24.5
26.67
25.7
21
25.4
23
26
25
23.5
22
25.5
27.3
24
21.59
21.59
28
24
24.6
25.4
25
26
23.6
26.5
27.5
21.59
24.1
27.94
24.1
24
27
24
24
29
26.2
25.4
24
23.9
23
24.3
22.5
22.5
25
24.13
25
24
23.5
24.765
22
24
22.8
28
""
25
23.8
Last Name Length
5
13
6
7
6
6
5
5
4
4
7
9
6
8
8
6
7
5
4
4
9
6
6
9
5
6
6
8
15
5
8
9
5
7
8
6
7
8
8
4
6
4
9
7
7
6
6
7
5
5
5
11
7
6
12
Best Pet
Cats
Dogs
Cats
Dogs
Dogs
Cats
Dogs
Cats
Cats
Cats
Dogs
Dogs
Dogs
Dogs
Cats
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Cats
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Cats
Cats
Dogs
Cats
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Dogs
Integer
3
5
3
4
3
3
3
3
3
3
3
4
3
4
1
4
3
4
2
3
3
3
4
5
3
3
4
3
4
2
3
4
2
3
2
3
4
3
2
3
2
3
4
3
3
4
4
4
3
4
3
3
2
4
4
Explanation / Answer
1) Use a 5% significance level to test the claim the average online student got less than 8 hours of sleep last night.
Soln:
Mean (µ) = 7
Std Dev (s) = 1.56
n = 55
The null hypothesis, H0
H0: µ0 = 8 (Mean sleep hours of student is 8 hours last night)
The alternative hypothesis, H1
H1: µ0 < 8 (Mean sleep hours of student is less than 8 hours last night)
The test statistic (rounded to the nearest hundredth)
z = (µ - µ0)/{s/n1/2} = - 4.74
alpha = 0.05
and Zcritical = 1.96
The P value
From z table P-value (Z = -3.4) = 0.0003
Hence P-value (Z = -4.74) < 0.0003
Decision
At alpha = 0.05, Since our Z < - Zcritical, we reject the null hypothesis.
Conclusion
At alpha = 0.05, we have statistical evidence that mean sleep hours of student was less than 8 hours last night.
2) Use a 1% significance level to test the claim a majority of online students will call "heads" will given the option to call a coin toss.
Soln:
From data,
proportion of heads(p) = 34/55 = 0.62
n = 55
The null hypothesis, H0
H0: p0 = 0.5 (Proportion of students calling heads is equal to 0.5)
The alternative hypothesis, H1
H1: p0 > 0.5 (Proportion of students calling heads is more than 0.5)
The test statistic (rounded to the nearest hundredth)
z = (p - p0)/{ p0*(1- p0)/n}1/2 = 1.75
alpha = 0.01
and Zcritical = 2.58
The P value
From z table P-value (Z <= 1.75) = 0.9599
Hence P-value (Z > 1.75) = 1 - 0.9599 = 0.0401
Decision
At alpha = 0.01, Since our Z < Zcritical, we fail to reject the null hypothesis.
Conclusion
At alpha = 0.01, we conclude that Proportion of students calling heads is equal to 0.5
3) Use a 5% significance level to test the claim the mean foot length for online students is less than 25 cm.
Soln:
Mean (µ) = 24.61
Std Dev (s) = 1.82
n = 54 (since one value is blank)
The null hypothesis, H0
H0: µ0 = 25 (Mean foot length is equal to 25cms)
The alternative hypothesis, H1
H1: µ0 < 25 (Mean foot length is less than 25cms)
The test statistic (rounded to the nearest hundredth)
z = (µ - µ0)/{s/n1/2} = - 1.58
alpha = 0.05
and Zcritical = 1.96
The P value
From z table,
P-value (Z <= -1.58) =0.0571
Decision
At alpha = 0.05, Since our Z > - Zcritical, we fail to reject the null hypothesis.
Conclusion
At alpha = 0.05, we do not have enough evidence to reject the null hypothesis ie Mean foot length is equal to 25cms.
4) Use a 5% significance level to test the claim the mean number of letters in the last names of online students is different from 6 letters
Soln:
Mean (µ) = 6.78
Std Dev (s) = 2.25
n = 55
The null hypothesis, H0
H0: µ0 = 6 (Mean number of letters in the last names of online students is equal to 6 letters)
The alternative hypothesis, H1
H1: µ0 <> 6 (Mean number of letters in the last names of online students is different from 6 letters)
The test statistic (rounded to the nearest hundredth)
z = (µ - µ0)/{s/n1/2} = 2.58
alpha = 0.05
and Zcritical = 1.96
The P value
From z table,
P-value (Z <= 2.58) = 0.9951
Decision
At alpha = 0.05, Since our Z > Zcritical, we reject the null hypothesis.
Conclusion
At alpha = 0.05, we conclude that Mean number of letters in the last names of online students is different from 6 letters.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.