2.3

Module Two Assignment

Minitab Hypothesis Testing

Michael Keister

QSO 620-Six Sigma Quality Management

Southern New Hampshire University

March 17, 2016

Problem One

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster by using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α=0.05. Assume that the swim times for the 25-yard freestyle are normal (Dean & Illowsky, 2016).

Data:

Mean Time: (ẍ) 16 (given)

Standard Deviation (Ơ) .8 (given)

Sample Size (n): 15 (given)

Significance Level (x): .05 (given)

Null Hypothesis u=16.43

Alternative Hypothesis u < 16.43

Rule: Reject Null Hypothesis if X > P-Valve

Minitab © Calculations:

One-Sample Z

Test of μ = 16.43 vs < 16.43

The assumed standard deviation = 0.8

N Mean SE Mean 95% Upper Bound Z P

15 16.000 0.207 16.340 -2.08 0.019

P Value is .019

Since X (.05) is > than the P-valve (.019) the null hypothesis is rejected. We believe that Jeffery will swim the 25 meter free style faster than 16.43 seconds with the new goggles.

Problem Two

A college football coach thought that his players could bench press a mean weight of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3)215(3) 225(1) 241(2) 252(2) 265(2) 275(2) 313(2) 316(5) 338(2) 341(1) 345(2) 368(2) 385(1). (Source: data from Reuben Davis, Kraig Evans, and Scott Gunderson.)

Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds. (Dean & Illowsky, 2016).

Data:

Mean Weight: (ẍ) 286.2 (calculated with Minitab©)

Standard Deviation (Ơ) 55 (given)

Sample Size (n): 30 (given)

Significance Level (x): .025 (given)

Null Hypothesis: µ = 275

Alternative Hypothesis: u > 275

Rule: Reject Null Hypothesis if P-Valve < .025

Minitab © Calculations:

One-Sample Z

Test of μ = 275 vs > 275

The assumed standard deviation = 55

N Mean SE Mean 97.5% Lower Bound Z P

30 286.2 10.0 266.5 1.12 0.132

P valve is .132

Since P Valve of .132 > X valve of .025 do not reject the Null Hypothesis. There is not sufficient evidence to state that the true mean weight lifted by the football team is more than 275 pounds.

Problem Three

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 7271. He performs a hypothesis test using a 5% level of significance. The data are from a normal distribution (Dean & Illowsky, 2016).

Sample Size (n): 10 (given)

Significance Level (X): .05 (given)

Mean Score: 67 (calculated using Minitab ©)

The standard deviation is not given for this example. It is stated in the example that the data is from a normal distribution. Therefore we can run a test using a t-test versus a z-test. T-test is best applied as long as the variables are approximately normally distributed and if you do not know the populations standard deviation (“Difference Between Z-Test and T-Test, 2011”).

So for this problem a T-test will be run on Minitab©

Null Hypothesis U = 65

Alternative Hypothesis U > 65

Rule: Reject Null Hypothesis if X > P-Valve

Minitab © Calculations:

One-Sample T: C1

Test of μ = 65 vs > 65

Variable N Mean StDev SE Mean 95% Lower Bound T P

C1 10 67.00 3.20 1.01 65.15 1.98 0.040

Since .05 > than the P valve of .04 the Null Hypothesis is rejected. The average test score is greater than 65 just as the teacher had predicted.

Problem Four

Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50%. Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance (Dean & Illowsky, 2016).

Sample Size (n): 100 (given)

Significance Level (x): .01 (given)

This is basically a yes or no problem to solve. The bride is either older or younger than their grooms. So 53 brides said yes they are older than their grooms so the mean is expressed in a percentage. .53 (53/100) of the brides state they are older than their grooms. At the same time .47 of the brides are younger than their grooms. Therefore on minitab we will use the test of a single population proportion.

As the problem states “is the same or different from” indicates a two tailed test.

Null Hypothesis p=.50

Alternative Hypothesis p≠.50

Minitab © Calculations:

Test and CI for One Proportion

Test of p = 0.5 vs p ≠ 0.5

Sample X N Sample p 99% CI Z-Value P-Value

1 53 100 0.530000 (0.401441, 0.658559) 0.60 0.549

Using the normal approximation.

The P value is calculated on .549.

Since this a proportion test we will compare the x valve and the p valve. The X valve (.01) < the P value (.549). Therefore with a 99% confidence level we cannot reject the Null Hypothesis. The data does not show sufficient evidence that the first time brides that are younger than their grooms is different than 50%.

Problem Five

Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones (Dean & Illowsky, 2016).

Sample Size (n): 150 (given)

Significance Level (x): .05 This is assumed as it is not given in the problem.

Again this is a single population proportion test. The cell phone company believes there is a 30% chance the household has three phones . So the test will be:

Null Hypothesis p=.30

Alternative Hypothesis p ≠.30

43 of the 150 households surveyed do have thee cell phones (43/150, .29) and 107 households do not have three phones (107/150, .72)

Minitab © Calculations:

Test and CI for One Proportion

Test of p = 0.3 vs p ≠ 0.3

Sample X N Sample p 95% CI Z-Value P-Value

1 43 150 0.286667 (0.214300, 0.359033) -0.36 0.722

Using the normal approximation.

P Value is .722

Comparing the P Value of .722 X (.05) is less than the p value of .722).

X < P value. Therefore the null hypothesis cannot be rejected as there is not enough evidence based on this sample size that the percentages of households having thee phones is not 30%.

Sources