Module 5 Minitab Assignment Solution

Module 5 Minitab Assignment Answers

  1. A hat company states that the mean hat size for a male is at least 7.25. A random sample of 32 hat sizes has a mean of 7.15 and a standard deviation of 0.35.
    1. At α = 0.05, can you reject the company’s claim that the mean hat size for a male is at least 7.25? Be sure to state the p-value for this test.
    1. In the context of the problem, describe what a Type I error and a Type II error would mean.

Answer:

Given,

Sample Size = 32

Sample mean = 7.15

Standard deviation = 0.35

α = 0.05

2. Process:Open Minitab SoftwareClick on “Stat” then select “Basic Structure”, then click on “1-Sample Z” option.Select “Summarized Data” then put the values of sample size, sample mean, standard deviation and hypothesis mean.Click on “Options”, then select “Mean<Hypothesized Mean” and click on “OK”The result will be shown on Session Window.

Descriptive Statistics

NMeanSE Mean95% Upper Bound
for μ
327.15000.06197.2518

μ: mean of Sample
Known standard deviation = 0.35

Test

Null hypothesisH₀: μ = 7.25
Alternative hypothesisH₁: μ < 7.25
Z-ValueP-Value 
-1.620.053 

So, the value of P is 0.053. At α = 0.05, P-value > α.

Null hypothesis is not rejected, and we can conclude that a hat company claims that the mean hat size for a male is 7.25.

b.

Type Error I:

If  P – value ≤ α ( = 0.05) reject the company’s claim that the mean hat size for a male is at least 7.25

Type Error II:

If  P – value > α ( = 0.05) accept the company’s claim that the mean hat size for a male is at least 7.25

2. The mean ACT score for 43 male high school students is 21.1 and the standard deviation is 5.0. The mean ACT score for 56 female high school students is 20.9 and the standard deviation is 4.7. At the 1% significance level, can you reject the claim that male and female high school students have equal ACT score averages? Be sure to use the p-value to make your conclusion.

Answer:

Let’s consider, Male as Sample 1 and Female as Sample 2

Given,

Sample 1 size = 43

Sample 2 size = 56

Sample 1 mean = 21.1 and standard deviation = 5.0

Sample 2 mean = 20.9 and standard deviation = 4.7

Process:

  • Open Minitab Software
    • Click on “Stat” then select “Basic Structure”, then click on “2-Sample T” option.
    • Select “Summarized Data” then put the values of sample size, sample mean and standard deviation of Sample 1 & 2.
    • Click on “Options”, then select “Difference<Hypothesized difference” as alternative hypothesis and click on “OK”
    • The result will be shown on Session Window.

Method

μ₁: mean of Sample 1
µ₂: mean of Sample 2
Difference: μ₁ – µ₂

Equal variances are assumed for this analysis.

Descriptive Statistics

SampleNMeanStDevSE Mean
Sample 14321.105.000.76
Sample 25620.904.700.63

Estimation for Difference

DifferencePooled
StDev
95% CI for
Difference
0.2004.832(-1.745, 2.145)

Test

Null hypothesisH₀: μ₁ – µ₂ = 0
Alternative hypothesisH₁: μ₁ – µ₂ ≠ 0
T-ValueDFP-Value 
0.20970.839 

Conclusion:

So, the value of  P = 0.839 , P > α(=0.01)

At the 1% significance level(α = 0.01),  Null hypothesis is not rejected, the claim that male and female high school students have equal ACT score averages

3. A medical research team conducted a study to test the effect of a migraine drug. Of the 400 subjects who took the drug, 100 were pain free after two hours. Of the 407 subjects that took a placebo, 40 were pain free after two hours. Can you reject the claim that the proportion of subjects who are pain-free is the same for the two groups? Use the appropriate hypothesis test at the 2% significance level. Be sure to use the p-value to make your conclusion.

Answer:

Let’s consider, Drug taker as Sample 1 and Placebo taker as Sample 2

Given,

Sample 1 size = 400 and event = 100

Sample 2 size = 407 and event = 40

Process:

  • Open Minitab Software
    • Click on “Stat” then select “Basic Structure”, then click on “Two Sample Proportion” option.
    • Select “Summarized Data” then put the values of number of events and trials of Sample 1 & 2.
    • Click on “Options”, then select “Difference<Hypothesized difference” as alternative hypothesis and make test method “Estimate the proportions separately” and click on “OK”
    • The result will be shown on Session Window.

Method

p₁: proportion where Sample 1 = Event
p₂: proportion where Sample 2 = Event
Difference: p₁ – p₂

Descriptive Statistics

SampleNEventSample p
Sample 14001000.250000
Sample 2407400.098280

Estimation for Difference

Difference95% CI for
Difference
0.151720(0.100367, 0.203073)

CI based on normal approximation

Test

Null hypothesisH₀: p₁ – p₂ = 0 
Alternative hypothesisH₁: p₁ – p₂ ≠ 0 
MethodZ-ValueP-Value
Normal approximation5.790.000
Fisher’s exact 0.000

Conclusion:

So, the value of P = 0.00, P < α(=0.02)

At the 2% significance level, Null hypothesis is rejecting the claim that the proportion of subjects who are pain-free is the same for the two groups.

4. The Federal Bureau of Prisons publishes data in Prison Statistics on the times served by prisoners released from federal institutions for the first time. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the following information on time served, in months. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses?

Dataset: 1

FraudFirearm
11.920.1
12.310.9
13.118.4
11.428.2
11.517.5
1412.2
11.218.9
7.719.3
9.320.4
12.814.8

Dataset: 2

FraudFirearm
9.218.9
9.616.6
10.522.1
10.719.4
10.220
11.521.4
12.312.5
12.213.4
13.919.5
9.910.8
  
  

Answer:

Here two data sets are given of Prison Statistics, both of them contained two samples : i. Fraud ii. Firearm

To determine the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses, we will use Paired T-Test method.

Process:

  • Open Minitab Software
    • Write the table details of Dataset 1 in the Worksheet For Dataset 1 ( Write the table details of Dataset 2 in the Worksheet For Dataset 2)
    • Click on “Stat” then select “Basic Structure”, then click on “Paired T-test” option.
    • Select “Summarized Data” then put the values of number of events and trials of Sample 1 & 2.
    • Click on “Options”, then select “Difference<Hypothesized difference” as alternative hypothesis and make test method “Estimate the proportions separately” and click on “OK”
    • The result will be shown on Session Window.

Results of Dataset 1:

Descriptive Statistics

SampleNMeanStDevSE Mean
Fraud1011.541.870.59
Firearm1018.154.891.55

Estimation for Paired Difference

MeanStDevSE Mean95% Upper Bound
for μ_difference
-6.615.911.87-3.18

µ_difference: mean of (Fraud – Firearm)

Test

Null hypothesisH₀: μ_difference = 0
Alternative hypothesisH₁: μ_difference < 0
T-ValueP-Value 
-3.540.003 

Conclusion:

So P – value( = 0.003) < α ( = 0.05) , reject the null hypothesis at the 5% significance level and conclude that the mean time served for fraud is less than that for firearms offenses

Results of Dataset 2:

Descriptive Statistics

SampleNMeanStDevSE Mean
Fraud1011.001.460.46
Firearm1017.463.941.25

Estimation for Paired Difference

MeanStDevSE Mean95% Upper Bound
for μ_difference
-6.464.271.35-3.99

µ_difference: mean of (Fraud – Firearm)

Test

Null hypothesisH₀: μ_difference = 0
Alternative hypothesisH₁: μ_difference < 0
T-ValueP-Value 
-4.790.000 

Conclusion:

So P – value( = 0.000) < α ( = 0.05) , reject the null hypothesis at the 5% significance level and conclude that the mean time served for fraud is less than that for firearms offenses

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