## 9.5 Additional Information and Full Hypothesis Test Examples

• 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 1

Ho: μ = 5
Ha: μ < 5
Significance level = 5%
Assume the p-value is 0.0243.

1. What type of test is this?
2. Determine if the test is left, right, or two-tailed.
3. Draw the picture of the p-value.
4. Do we reject null hypothesis, H0: μ = 5?
5. Do we have enough evidence to conclude that μ < 5?

### Try It

H0: μ = 10
Ha: μ < 10
Significance level = 5% = 0.05
Assume the p-value is 0.0435.

1. What type of test is this?
2. Determine if the test is left, right, or two-tailed.
3. Draw the picture of the p-value.
4. Do we reject null hypothesis, H0: μ = 10?
5. Do we have enough evidence to conclude that μ < 10?

## Example 2

H0: p ≤ 0.2
Ha: p > 0.2
Significance level = 0.05
Assume the p-value is 0.0719.

1. What type of test is this?
2. Determine if the test is left, right, or two-tailed.
3. Draw the picture of the p-value.
4. Do we reject null hypothesis, H0p ≤ 0.2?
5. Do we have enough evidence to conclude that p > 0.2?

### Try It

H0: μ ≤ 1
Ha: μ > 1
Significance level = 1%
Assume the p-value is 0.1243.

1. What type of test is this?
2. Determine if the test is left, right, or two-tailed.
3. Draw the picture of the p-value.
4. Do we reject null hypothesis, H0: μ ≤ 1?
5. Do we have enough evidence to conclude that μ > 1?

## Example 3

H0: p = 50
Ha: p ≠ 50
Significance level = 1%
Assume the p-value is 0.0005
1. What type of test is this?
2. Determine if the test is left, right, or two-tailed.
3. Draw the picture of the p-value.
4. Do we reject null hypothesis, H0: p = 50?
5. Do we have enough evidence to conclude that p ≠ 50?

### Try It

H0: p = 0.5
Ha: p ≠ 0.5
Significance level = 0.05
Assume the p-value is 0.2564.

1. What type of test is this?
2. Determine if the test is left, right, or two-tailed.
3. Draw the picture of the p-value.
4. Do we reject null hypothesis, H0: p = 0.5?
5. Do we have enough evidence to conclude that p ≠ 0.5?

## Steps to set up a hypothesis test:

• Set up H0 and Ha.
• Determine the significance level.
• Find p-value.
• Compare p-value and significance level, ($\alpha$.
• Decide if we reject / not reject H0.
 If p-value < significance level, then reject  H0. Therefore, enough evidence to conclude Ha. If p-value > significance level, then not reject  H0. Therefore, not enough evidence to conclude Ha.
• Conclusion.

# 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 5% significance level. Assume that the swim times for the 25-yard freestyle are normal.

### Solution:

mean = 16.43 seconds, standard deviation = 0.8 seconds.

Since the problem is about a mean, this is a test of a single population mean.

#### Using TI-Calculator to find p-value:

• 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 ) and 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:

To find P$\left(\overline{x}<{16}\right)$, we will use TI-Calculator.
Ti-Calculator: 2nd DISTR normcdf (${-10}^{99}$, 16,16.43,$\frac{{0.8}}{{\sqrt{15}}}$).

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.)

### 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 2 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.

## Example 4

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:

 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 5

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. The data are assumed to be from a normal distribution.
He performs a hypothesis test using a 5% level of significance.

### The p-value can easily be calculated.

• Put the data into a list.
• Press STAT and arrow over to TESTS .
• Press 2:T-Test . (as we do not have the population standard deviation. )
• 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 . A
• 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

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.

## Example 6

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 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.

### Solution:

• 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.

### Try It

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 7

Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company survey 150 households with the result that 43 of the households have three cell phones. They believe that the proportion is less than 30%. Conduct a hypothesis test to check their claim at 5% significance level.

### 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.

## Example 8

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. (Assume the population is normal. )

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 1?
Use a significance level of 0.05.

### Try It

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 0.5% significance level to run the hypothesis test.

## Example 9

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 0.07734%. 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 1% significance level.

# Concept Review

The hypothesis test itself has an established process. This can be summarized as follows:

Determine H0 and Ha. Remember, they are contradictory.
Determine the random variable.
Determine the distribution for the test.
Draw a graph, calculate the test statistic, and use the test statistic to calculate the p-value. (A z-score and at-score are examples of test statistics.)
Compare the preconceived α with the p-value, make a decision (reject or do not reject H0), and write a clear conclusion using English sentences.

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.