Learning Outcomes
- Conduct and interpret hypothesis tests for a single population proportion.
- In a hypothesis test problem, you may see words such as “the level of significance is 1%.” The “1%” is the preconceived or preset α.
- The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
- If no level of significance is given, a common standard to use is α = 0.05.
- When you calculate the p-value and draw the picture, the p-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
- The alternative hypothesis, Ha, tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
- Ha never has a symbol that contains an equal sign.
- Thinking about the meaning of the p-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller p-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value such as 0.4, as opposed to a p-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.
The following examples illustrate a left-, right-, and two-tailed test.
Example
Ho: μ = 5, Ha: μ < 5
Test of a single population mean. Ha tells you the test is left-tailed. The picture of the p-value is as follows:
TRY IT
H0: μ = 10, Ha: μ < 10
Assume the p-value is 0.0935. What type of test is this? Draw the picture of the p-value.
Solution:
left-tailed test
Example
H0: p ≤ 0.2 Ha: p > 0.2
This is a test of a single population proportion. Ha tells you the test is right-tailed. The picture of the p-value is as follows:
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H0: μ ≤ 1, Ha: μ > 1
Assume the p-value is 0.1243. What type of test is this? Draw the picture of the p-value.
Solution:
right-tailed test
Example
This is a test of a single population mean. Ha tells you the test is two-tailed. The picture of the p-value is as follows.
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H0: p = 0.5, Ha: p ≠ 0.5
Assume the p-value is 0.2564. What type of test is this? Draw the picture of the p-value.
Solution:
two-tailed test
Full Hypothesis Test Examples
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 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.
Solution
Set up the Hypothesis Test:
Since the problem is about a mean, this is a test of a single population mean.
H0: μ = 16.43 Ha: μ < 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: [latex]\overline{X}[/latex]= the mean time to swim the 25-yard freestyle.
Distribution for the test:[latex]\overline{X}[/latex] is normal (population standard deviation is known: σ = 0.8)
μ = 16.43 comes from H0 and not the data. σ = 0.8, and n = 15.
Calculate the p-value using the normal distribution for a mean:
p-value = P[latex]\left(\overline{x}<{16}\right)[/latex]= 0.0187 where the sample mean in the problem is given as 16.
p-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 = 16.43 comes from H0. Our assumption is μ = 16.43.
Interpretation of the p-value: If H0 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 α and the p-value:
α = 0.05 p-value = 0.0187 α > p-value
Make a decision: Since α > p-value, reject H0.
This means that you reject μ = 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.
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 μ0 (null hypothesis), .8 for σ, 16 for the sample mean, and 15 for n. Arrow down to μ : (alternate hypothesis) and arrow over to < μ0. Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p-value (p = 0.0187) but it also calculates the test statistic (z-score) for the sample mean. μ < 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with z = -2.08 (test statistic) and p = 0.0187 (p-value). Make sure when you use Draw that no other equations are highlighted in 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[latex]\left(\overline{x}<{16}\right)[/latex]2nd DISTR normcdf ([latex]{-10}^{99}[/latex], 16,16.43,[latex]\frac{{0.8}}{{\sqrt{15}}}[/latex]).
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.)
TRY IT
The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.
First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.
Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for μ0 (null hypothesis), 2 for σ, 45 for the sample mean, and 20 for n. Arrow down to μ: (alternative hypothesis) and set it either as <, ≠, or >. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value but it also calculates the test statistic (z-score) for the sample mean. Select <, ≠, or >; for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and p-value. Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.
Solution:
Since the problem is about a mean, this is a test of a single population mean.
H0 : μ = 40
Ha : μ > 40
p = 0.0062
Because p < α, we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.
The traditional way to compare the two probabilities, α and the p-value, is to compare the critical value (z-score from α) to the test statistic (z-score from data). The calculated test statistic for the p-value is –2.08. (From the Central Limit Theorem, the test statistic formula is z=[latex]\frac{\overline{x}-{\mu}_{X}}{(σ)}[/latex]. For this problem,[latex]\overline{x}[/latex] = 16,[latex]{/mu}_{X}[/latex] = 16.43 from the null hypothes is, [latex]{/mu}_{X}[/latex] = 0.8, and n = 15.) You can find the critical value for α = 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 (which demonstrates that α > p-value), reject H0. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities α and the p-value is very common. For this problem, the p-value, 0.0187 is considerably smaller than α, 0.05. You can be confident about your decision to reject. The graph shows α, the p-value, and the test statistics and the critical value.
Example
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).
Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds.
Solution:
Set up the Hypothesis Test:
Since the problem is about a mean weight, this is a test of a single population mean.
H0: μ = 275
Ha: μ > 275
This is a right-tailed test.
Calculating the distribution needed:
Random variable: [latex]\overline{X}[/latex] = the mean weight, in pounds, lifted by the football players.
Distribution for the test: It is normal because σ is known. [latex]\overline{X}~N\left(275,\frac{55}{\sqrt{30}}\right)[/latex]
[latex]\overline{x}[/latex] = 286.2
σ =55 pounds (Always use σ if you know it.) We assume μ = 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. p-value=P[latex]\left(\overline{x}>286.2\right)=0.1323[/latex].
Interpretation of the p-value: If H0 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 α and the p-value:
α = 0.025 p-value = 0.1323
Make a decision: Since α <p-value, do not reject H0.
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.
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 μ0, 55 for σ, 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: and arrow over to > μ0. Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p-value (p = 0.1331, a little different from the previous 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 > 275 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with z = 1.112 (test statistic) and p = 0.1331 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.
Example
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 assumed to be from a normal distribution.
Solution:
Set up the hypothesis test:
A 5% level of significance means that α = 0.05. This is a test of a single population mean.
H0: μ = 65 Ha: μ > 65
Since the instructor thinks the average score is higher, use a “>”. The “>” means the test is right-tailed.
Determine the distribution needed:
Random variable: [latex]\overline{X}[/latex] = 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 = 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 tdf. Therefore, the distribution for the test is t9 where n = 10 and df = 10 – 1 = 9.
Calculate the p-value using the Student’s t-distribution:
p-value = P([latex]\overline{x}[/latex]> 67) = 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 65 or more.
Compare α and the p-value:
Since α = 0.05 and p-value = 0.0396. α > p-value.
Make a decision: Since α > p-value, reject H0.
This means you reject μ = 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.
TRY IT
It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors.
Solution:
H0: μ = 5
Ha: μ < 5
p = 0.0082
Because p < α, we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.
Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).
Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).
Example
Set up the hypothesis test:
The 1% level of significance means that α = 0.01. This is a test of a single population proportion.
H0: p = 0.50 Ha: p ≠ 0.50
The words “is the same or different from” tell you this is a two-tailed test.
Calculate the distribution needed:
Random variable: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′, the estimated proportion.
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A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.
First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.
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 not 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.
try it
Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p-value, state your conclusion, and identify the Type I and Type II errors.
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 p. The distribution for the test is normal. The estimated proportion p′ is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α = 0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!
My dog has so many fleas,
They do not come off with ease.
As for shampoo, I have tried many types
Even one called Bubble Hype,
Which only killed 25% of the fleas,
Unfortunately I was not pleased.
I’ve used all kinds of soap,
Until I had given up hope
Until one day I saw
An ad that put me in awe.
A shampoo used for dogs
Called GOOD ENOUGH to Clean a Hog
Guaranteed to kill more fleas.
I gave Fido a bath
And after doing the math
His number of fleas
Started dropping by 3’s!
Before his shampoo
I counted 42.
At the end of his bath,
I redid the math
And the new shampoo had killed 17 fleas.
So now I was pleased.
Now it is time for you to have some fun
With the level of significance being .01,
You must help me figure out
Use the new shampoo or go without?
The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.
1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95
Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.
In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.
According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.
Concept Review
The hypothesis test itself has an established process. This can be summarized as follows:
Notice that in performing the hypothesis test, you use α and not β. β is needed to help determine the sample size of the data that is used in calculating the p-value. Remember that the quantity 1 – β is called thePower of the Test. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.