Full Hypothesis Test Examples

Learning Outcomes

Conduct a hypothesis test for one proportion and interpret the conclusion in context

Example 4

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.

The first solution is step-by-step.

Show Solution A

The second solution uses the TI-83+ and TI-84 calculators.

Show Solution B

TRY IT 4

The mean throwing distance of a football for 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.

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HISTORICAL NOTE (EXAMPLE 1)

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]\dfrac{{\overline{x}}-{\mu_X}}{\left ( \dfrac{\sigma_X}{\sqrt n} \right )}[/latex].

For this problem, [latex]\overline{x}[/latex] = 16, [latex]{\mu}_{X}[/latex] = 16.43 from the null hypothesis, [latex]{\sigma}_{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, the test statistic, and the critical value.

Distribution curve comparing the α to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. α is equal to 0.05 and points to the area between the values of -2.15 and -1.645.

Example 5

A college football coach records the mean weight that his players can bench press as 275 pounds with a standard deviation of 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.

Show Answer

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
H_a: 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}[/latex]~[latex]N\left(275,\frac{55}{\sqrt{30}}\right)[/latex]

[latex]\overline{x}[/latex] = 286.2 (from the data).

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 (see Table F2 for using the data as input):

p-value = P([latex]\overline{x}[/latex] > 286.2) = 0.1323.

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.

Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.

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 6

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.

Show Answer

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 H_a: 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 t_df. 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.

Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.

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.

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 μ₀, the name of the list where you put the data, and 1 for Freq:.
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 = 0.0396) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation.
μ > 65 is the alternative hypothesis.
Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER.
A shaded graph appears with t = 1.9781 (test statistic) and p = 0.0396 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.

TRY IT 5

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.

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Example 7

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.

Show Answer

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 H_a: 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.

P’ ~ N[latex]\left(p,\sqrt{\dfrac{{p\cdot{q}}}{{n}}}\right)[/latex] Therefore P’ ~N[latex]\left(0.5,\sqrt{\dfrac{{0.5\cdot{0.5}}}{{100}}}\right)[/latex]

where p = 0.50, q = 1 – p = 0.50, and n = 100

Calculate the p-value using the normal distribution for proportions:

p-value = P (p′ < 0.47 or p′ > 0.53) = 0.5485

where x = 53, p′ = [latex]\dfrac{x}{n}[/latex] = [latex][/latex]\dfrac{53}{100} = 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’ is 0.53 or more OR 0.47 or less (see the graph below).

Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.

μ = p = 0.50 comes from H₀, the null hypothesis.

p′ = 0.53. Since the curve is symmetrical and the test is two-tailed, the p′ for the left tail is equal to 0.50 – 0.03 = 0.47 where μ = p = 0.50. (0.03 is the difference between 0.53 and 0.50.)

Compare α and the p-value:

Since α = 0.01 and p-value = 0.5485. α < p-value.

Make a decision: Since α < p-value, you cannot reject H₀.

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.

The p-value can easily be calculated.

Press STAT and arrow over to TESTS.
Press 5:1-PropZTest.
Enter .5 for p0, 53 for x and 100 for n.
Arrow down to Prop and arrow to not equals p₀. Press ENTER.
Arrow down to Calculate and press ENTER.
The calculator calculates the p-value (p = 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 = 0.6 (test statistic) and p = 0.5485 (p-value). Make sure when you use Draw that no other equations are highlighted in 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 who 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 who 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.)

TRY IT 6

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.

Example 8

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.

1. The value that helps determine the p-value is p’. Calculate p’.

2. What is a success for this problem?

3. What is the level of significance?

4. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately. Calculate the p-value.

5. Make a decision. _____________(Reject/Do not reject) H₀ because____________.

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try it 7

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!

Example 9

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?

Show Answer

Example 10

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.

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Example 11

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.

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Example 12

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.

Show Answer

alpha	decision	reason for decision
0.01	Do not reject H₀	a < p-value

Module 9: Hypothesis Testing With One Sample

Learning Outcomes

Example 4

TRY IT 4

USING THE TI-83, 83+, 84, 84+ CALCULATOR

Example 5

Example 6

TRY IT 5

Example 7

TRY IT 6

Example 8

try it 7

Example 9

Example 10

Example 11

Example 12