Assignment: hypothesis testing using minitab

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Hypothesis Testing of Single Mean and Single Proportion: ExamplesPage by: Susan Dean,Barbara Illowsky, Ph.D.Books
SummaryJeffrey, 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=0.05α=0.05. Assume that the swim times for the 25-yard freestyle are normal.
Set up the Hypothesis Test:
Since the problem is about a mean, this is a test of a single population mean.
HoHo: μ=16.43μ=16.43 HaHa: μμ <16.43<16.43
For Jeffrey to swim faster, his time will be less than 16.43 seconds. The “<” tells you this is left-tailed.
Determine the distribution needed:
Random variable: X¯¯¯X = the mean time to swim the 25-yard freestyle.
Distribution for the test: X¯¯¯X is normal (population standard deviation is known: σσ =0.8=0.8)
X¯¯¯X ~ N(μ,σXn√)N(μ,σXn) Therefore, X¯¯¯X ~ NN (16.43,0.815√)(16.43,0.815)μ=16.43μ=16.43 comes from H0H0 and not the data. σ=0.8σ=0.8, and n=15n=15.
Calculate the p-value using the normal distribution for a mean:
p-value=P(x−<16)=0.0187p-value=P(x-<16)=0.0187 where the sample mean in the problem is given as 16.
p-value=0.0187p-value=0.0187 (This is called the actual level of significance.) The p-value is the area to the left of the sample mean is given as 16.
Graph:
μ=16.43μ=16.43 comes from HoHo. Our assumption is μ=16.43μ=16.43.
Interpretation of the p-value: If HoHo is true, there is a 0.0187 probability (1.87%) that Jeffrey’s mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.
Compare αα and the p-value:
α=0.05p-value=0.0187α>p-valueα=0.05p-value=0.0187α>p-valueMake a decision: Since α>p-valueα>p-value, reject HoHo.
This means that you reject μ=16.43μ=16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles.
Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey’s mean time to swim the 25-yard freestyle is less than 16.43 seconds.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 16.43 for μ0μ0 (null hypothesis), .8 for σσ, 16 for the sample mean, and 15 for nn. Arrow down to μμ: (alternate hypothesis) and arrow over to <μ0μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p=0.0187p=0.0187) but it also calculates the test statistic (z-score) for the sample mean. μ<16.43μ16.43 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z=−2.08z=-2.08 (test statistic) and p=0.0187p=0.0187 (p-value). Make sure when you use Draw that no other equations are highlighted in Y=Y= and the plots are turned off.
When the calculator does a Z-Test, the Z-Test function finds the p-value by doing a normal probability calculation using the Central Limit Theorem:
P(x¯P(x < 1616 ) == 2nd DISTR normcdf (−10^99,16,16.43,0.8/15−−√)(-10^99,16,16.43,0.8/15).
The Type I and Type II errors for this problem are as follows:
The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)
The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)
Historical Note: The traditional way to compare the two probabilities, αα and the p-value, is to compare the critical value (z-score from αα) to the test statistic (z-score from data). The calculated test statistic for the p-value is −2.08-2.08. (From the Central Limit Theorem, the test statistic formula is z=x¯−μX(σXn√)z=x-μX(σXn). For this problem, x¯=16x=16, μX=16.43μX=16.43 from the null hypothesis, σX=0.8σX=0.8, and n=15n=15.) You can find the critical value for α=0.05α=0.05 in the normal table (see 15.Tables in the Table of Contents). The z-score for an area to the left equal to 0.05 is midway between -1.65 and -1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is -1.645. Since −1.645>−2.08-1.645>-2.08 (which demonstrates that α>p-valueα>p-value), reject HoHo. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities αα and the p-value is very common. For this problem, the p-value, 0.01870.0187 is considerably smaller than αα, 0.05. You can be confident about your decision to reject. The graph shows αα, the p-value, and the test statistics and the critical value.
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.
Set up the Hypothesis Test:
Since the problem is about a mean weight, this is a test of a single population mean.
HoHo: μμ =275=275 HaHa: μμ >275>275 This is a right-tailed test.
Calculating the distribution needed:
Random variable: X¯¯¯X = the mean weight, in pounds, lifted by the football players.
Distribution for the test: It is normal because σσ is known.
X¯¯¯X ~ NN (275,5530√)(275,5530)x¯x =286.2=286.2 pounds (from the data).
σ=55σ=55 pounds (Always use σσ if you know it.) We assume μ=275μ=275 pounds unless our data shows us otherwise.
Calculate the p-value using the normal distribution for a mean and using the sample mean as input (see the calculator instructions below for using the data as input):
p-value=P(p-value=P( x¯x >286.2>286.2) =0.1323=0.1323.
Interpretation of the p-value: If HoHo is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.
Compare αα and the p-value:
α=0.025p-value=0.1323α=0.025p-value=0.1323
Make a decision: Since αα<p-valuep-value, do not reject HoHo.
Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Put the data and frequencies into lists. Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Data and press ENTER. Arrow down and enter 275 for μ0μ0, 55 for σσ, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to μ:μ: and arrow over to >μ0>μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p=0.1331p=0.1331, a little different from the above calculation – in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation. μ>275μ>275 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z=1.112z=1.112 (test statistic) and p=0.1331p=0.1331 (p-value). Make sure when you use Draw that no other equations are highlighted in Y=Y= and the plots are turned off.
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 72 71. He performs a hypothesis test using a 5% level of significance. The data are from a normal distribution.
Set up the Hypothesis Test:
A 5% level of significance means that α=0.05α=0.05. This is a test of a single population mean.
HoHo: μμ =65=65 HaHa: μμ >65>65Since the instructor thinks the average score is higher, use a “>>”. The “>>” means the test is right-tailed.
Determine the distribution needed:
Random variable: X¯¯¯X = average score on the first statistics test.
Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given n=10n=10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student’s-t.
Use tdftdf. Therefore, the distribution for the test is t9t9 where n=10n=10 and df=10−1=9df=10-1=9.
Calculate the p-value using the Student’s-t distribution:
p-value=P(p-value=P( x¯x >67>67 )=0.0396=0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.
Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 67 or more.
Compare αα and the p-value:
Since α=.05α=.05 and p-value=0.0396p-value=0.0396. Therefore, α>p-valueα>p-value.
Make a decision: Since α>p-valueα>p-value, reject HoHo.
This means you reject μ=65μ=65. In other words, you believe the average test score is more than 65.
Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Put the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 65 for μ0μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ:μ: and arrow over to >μ0>μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p=0.0396p=0.0396) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation. μ>65μ>65 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with t=1.9781t=1.9781 (test statistic) and p=0.0396p=0.0396 (p-value). Make sure when you use Draw that no other equations are highlighted in Y=Y= and the plots are turned off.
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.
Set up the Hypothesis Test:
The 1% level of significance means that α=0.01α=0.01. This is a test of a single population proportion.
HoHo: pp =0.50=0.50 HaHa: pp ≠0.50≠0.50The words “is the same or different from” tell you this is a two-tailed test.
Calculate the distribution needed:
Random variable: P’P’ = the percent of of first-time brides who are younger than their grooms.
Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P’P’, the estimated proportion.
P’P’ ~ NN (p,p⋅qn−−−√)(p,p⋅qn) Therefore, P’P’ ~ NN (0.5,0.5⋅0.5100−−−−−√)(0.5,0.5⋅0.5100) where p=0.50p=0.50, q=1−p=0.50q=1-p=0.50, and n=100n=100.
Calculate the p-value using the normal distribution for proportions:
p-value=P(p’p-value=P(p'< 0.470.47 or p’>0.53p’>0.53 ) =0.5485=0.5485
where x=53x=53, p’=xnp’=xn =53100=0.53=53100=0.53.
Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion p’p’ is 0.53 or more OR 0.47 or less (see the graph below).
μ=p=0.50μ=p=0.50 comes from HoHo, the null hypothesis.
p’p’=0.53=0.53. Since the curve is symmetrical and the test is two-tailed, the p’p’ for the left tail is equal to 0.50−0.03=0.470.50-0.03=0.47 where μ=p=0.50μ=p=0.50. (0.03 is the difference between 0.53 and 0.50.)
Compare αα and the p-value:
Since α=0.01α=0.01 and p-value=0.5485p-value=0.5485. Therefore, αα< p-valuep-value.
Make a decision: Since αα<p-valuep-value, you cannot reject HoHo.
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides that are younger than their grooms is different from 50%.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Press STAT and arrow over to TESTS. Press 5:1-PropZTest. Enter .5 for p0p0, 53 for xx and 100 for nn. Arrow down to Prop and arrow to not equals p0p0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator calculates the p-value (p=0.5485p=0.5485) and the test statistic (z-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z=0.6z=0.6 (test statistic) and p=0.5485p=0.5485 (p-value). Make sure when you use Draw that no other equations are highlighted in Y=Y= and the plots are turned off.
The Type I and Type II errors are as follows:
The Type I error is to conclude that the proportion of first-time brides that are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true).
The Type II error is there is not enough evidence to conclude that the proportion of first time brides that are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)
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.
Set up the Hypothesis Test:
HoHo: pp =0.30=0.30 HaHa: pp≠ 0.300.30Determine the distribution needed:
The random variable is P’P’ = proportion of households that have three cell phones.
The distribution for the hypothesis test is P’P’ ~ NN (0.30,(0.30)⋅(0.70)150−−−−−−−−√)(0.30,(0.30)⋅(0.70)150)
The value that helps determine the p-value is p’p’. Calculate p’p’.
p’=xnp’=xn where xx is the number of successes and nn is the total number in the sample.
x=43x=43, n=150n= 150p’ = 4315043150What is a success for this problem?
A success is having three cell phones in a household.
What is the level of significance?
The level of significance is the preset αα. Since αα is not given, assume that α=0.05α=0.05.
Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.
Calculate the p-value.
p-value = 0.7216
Make a decision. _____________(Reject/Do not reject) H0H0 because____________.
Assuming that αα = 0.05, α<p-valueα < p-value. The Decision is do not reject H0H0 because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.
The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter pp. The distribution for the test is normal. The estimated proportion p’p’ is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α=0.01α=0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!

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