8.6 Using Technology for Hypothesis Testing with One Sample

Jared Eusea; Phyllis Okwan; Rachid Belmasrour; Stephan Patterson; Stephen Andrus

Chapter 8: Hypothesis Testing with One Sample

8.6 Using Technology for Hypothesis Testing with One Sample

Learning Objectives

By the end of this section, the student should be able to:

use technology to solve problems involving Hypothesis Testing with One Sample

This is another section to show technology tools that can be used to help with the statistics concepts taught in this chapter. This section will cover tools to help solve problems that are Hypothesis Testing with One Sample.

Graphing Calculator

Similar to all the other technology sections, the main tool introduced will be the TI-83, 83+, 84, or 84+ Graphing Calculator. The graphing calculator can help to perform the entire hypothesis test for one sample, including getting the test statistic and p-value. It can also get a graph of the shaded p-value region, if you'd like that.

**Be aware!! The graphing calculator cannot determine the hypothesis for the test or which test you need to do, and it will not help you make the correct conclusion for the test. It only takes what you input into it.

Let's look at the options we have in the graphing calculator for hypothesis testing with one sample:

Hypothesis Testing with One Sample in the TI-83, 83+, 84, or 84+ Graphing Calculator

Access statistics mode by pressing .
Navigate to <TESTS> by using the .

On this screen you will find all of the available hypothesis tests for on sample:

<Z-Test> is the hypothesis test for a single mean when σ is known.
<T-Test> is the hypothesis test for a single mean when σ is unknown; s estimates σ.
<1-PropZTest> is the hypothesis test for a single proportion.

Highlight your choice and press enter key .

Notes:

You can enter the data into a list and use <DATA> to have the calculator find the sample mean and standard deviation, or you may enter the sample mean and standard deviation directly by using <STATS>.
The null hypothesis value is input in the row below <Inpt>. For a test of a single mean <μ0> represents the null hypothesis. For a test of a single proportion, <p∅> represents the null hypothesis.
Enter the alternate hypothesis on the bottom row by highlighting the appropriate tailed test after <μ> or <prop>.
Selecting <Calculate> will show you the value of the test statistic, p-value, the sample mean, and sample standard deviation.
Selecting <Draw> will show you the shaded region under the normal or Student's t curve with the z- or t-score and the p-value. Make sure when you use <Draw> that no other equations are highlighted in and the plots are turned off ( , [STAT PLOT]).

Legend

represents a button press
[ ] represents yellow command or green letter behind a key
< > represents items on the screen

Let's now look at specific examples of each.

Hypothesis Tests for a Single Population Mean, Population Standard Deviation Known

For the examples below, the by-hand method will still be shown as it did in 8.5 Additional Information and Full Hypothesis Test Examples, because doing hypothesis tests is a lot more than just plugging into the calculator - you need to know the process involved. You will see a box that separates the calculator information.

Example

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.

[latex]{H}_{0}: \mu = 16.43, {H}_{a}: \mu \lt 16.43[/latex]

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: [latex]\sigma = 0.8[/latex])

[latex]\overline{X} \sim N\left(\mu ,\frac{{\sigma }_{X}}{\sqrt{n}}\right)[/latex]

Therefore, [latex]\overline{X} \sim N\left(16.43,\frac{0.8}{\sqrt{15}}\right)[/latex]

[latex]\mu = 16.43[/latex] comes from [latex]{H}_{0}[/latex] and not the data. [latex]\sigma = 0.8[/latex], and [latex]n = 15[/latex].

Calculate the p-value using the normal distribution for a mean:

[latex]\text{p-value} = 0.0187[/latex] 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:

A normal distribution with 16.43 in the middle. A value of 16 with a shaded region shaded to the left of it as the p-value. — **Figure 1.** Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A shaded region in the left tail of the curve.

[latex]\mu = 16.43[/latex] comes from[latex]{H}_{0}[/latex]. Our assumption is [latex]\mu = 16.43[/latex].

Interpretation of the p-value: If [latex]{H}_{0}[/latex] 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 [latex]\alpha[/latex] and the p-value: [latex]\alpha = 0.05[/latex] and [latex]\text{p-value} = 0.0187[/latex], so [latex]\alpha > \text{p-value}[/latex]

Make a decision: Since [latex]\alpha > \text{p-value}[/latex], reject [latex]H_{0}[/latex].

This means that you reject [latex]\mu = 16.43[/latex]. 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 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 no 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.)

Using the TI-83, 83+, 84, 84+ Calculator

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 [latex]\mu 0[/latex] (null hypothesis), .8 for [latex]\sigma[/latex], 16 for the sample mean, and 15 for[latex]n[/latex]. Arrow down to [latex]\mu[/latex]: (alternate hypothesis) and arrow over to [latex]\lt \mu 0[/latex]. 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 [latex]z = -2.08[/latex] (test statistic) and [latex]p = 0.0187[/latex] (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.

Note

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: [latex]P\left(\overline{x} \lt 16 \right)= \text{2nd DISTR normcdf} \left(-10^{99},16,16.43,0.8/\sqrt{15}\right)[/latex]

Your Turn!

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.

Solution

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

[latex]H_{0} : \mu = 40[/latex]

[latex]H_{a} : \mu > 40[/latex]

[latex]p = 0.0062[/latex]

A normal distribution with 40 in the middle. A value of 45 with a shaded region shaded to the right of it as the p-value. — **Figure 2.** Normal distribution curve with values 40, as the population mean, and 45, as the point to determine the p-value

Because [latex]p \lt \alpha[/latex], we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.

Using the TI-83, 83+, 84, 84+ Calculator

Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for [latex]\mu 0[/latex] (null hypothesis), 2 for [latex]\sigma[/latex], 45 for the sample mean, and 20 for [latex]n[/latex]. Arrow down to [latex]\mu[/latex]: (alternative hypothesis) and set it either as [latex]\lt, \neq, \text{ or } >[/latex]. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value, [latex]p = 0.0062[/latex], but it also calculates the test statistic (z-score) for the sample mean. Because [latex]p \lt \alpha[/latex], we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistics and p-value. Make sure when you use Draw that no other equations are highlighted in [latex]Y =[/latex] and the plots are turned off.

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); and 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.

[latex]H_{0}: \mu = 275, H_{a}: \mu > 275[/latex]

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 [latex]\sigma[/latex] is known.

[latex]\overline{X} \sim N\left(275,\frac{55}{\sqrt{30}}\right)[/latex]

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

[latex]\sigma = 55 \text{ pounds}[/latex] (Always use [latex]\sigma[/latex] if you know it.) We assume [latex]\mu = 275 \text{ pounds}[/latex] unless our data shows us otherwise.

[latex]p\text{-value}= 0.1323[/latex].

Interpretation of the p-value: If [latex]H_0[/latex] 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.

A normal distribution with 275 in the middle. A value of 286.2 with a shaded region shaded to the right of it as p-value = 0.1323 — **Figure 3.** Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. The p-value points to the area to the right of 286.2.

Compare [latex]\alpha[/latex] and the p-value: [latex]\alpha = 0.025, \text{ p-value} = 0.1323[/latex]

Make a decision: Since [latex]\alpha \lt \text{p-value}[/latex], do not reject [latex]H_{0}[/latex].

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.

Using the TI-83, 83+, 84, 84+ Calculator

To use the calculator, 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 [latex]\mu 0[/latex], 55 for [latex]\sigma[/latex], the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to [latex]\mu[/latex]: and arrow over to [latex]> \mu 0[/latex]. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value ([latex]p = 0.1331[/latex] (a little different than if you did it by-hand because you would probably use the sample mean rounded to one decimal place instead of the data). It also calculates the test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation. [latex]\mu > 275[/latex] is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with [latex]z = 1.112[/latex] (test statistic) and [latex]p = 0.1331[/latex] (p-value). Make sure when you use Draw that no other equations are highlighted in [latex]Y =[/latex] and the plots are turned off.

Hypothesis Tests for a Single Population Mean, Population Standard Deviation Unknown

Even though we are now shifting to hypothesis tests with an unknown population standard deviation, again, the by-hand method will still be shown as it did in 8.5 Additional Information and Full Hypothesis Test Examples. The same thing as before, doing hypothesis tests is a lot more than just plugging into the calculator - you need to know the process involved. You will see a box that separates the calculator information.

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, and 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 [latex]\alpha = 0.05[/latex]. This is a test of a single population mean.

[latex]H_{0}: \mu = 65, H_{a}: \mu > 65[/latex]

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 [latex]n = 10[/latex] 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 [latex]n = 10[/latex] and [latex]df = 10 - 1 = 9[/latex].

Calculate the p-value using the Student's t-distribution:

[latex]\text{p-value} = 0.0396[/latex] 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.

A normal distribution with 65 in the middle. A value of 67 with a shaded region shaded to the right of it as p-value = 0.0396 — **Figure 4.** Normal distribution curve of the scores of the first statistics test with values of 65 and 67 on the x-axis. The p-value points to the area to the right of 67.

Compare [latex]\alpha[/latex] and the p-value:

Since [latex]\alpha = 0.05[/latex] and [latex]\text{p-value} = 0.0396[/latex], [latex]\alpha > \text{p-value}[/latex].

Make a decision: Since [latex]\alpha > \text{p-value}[/latex], reject [latex]H_{0}[/latex].

This means you reject [latex]\mu = 65[/latex]. 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, 83+, 84, 84+ Calculator

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 [latex]\mu 0[/latex], the name of the list where you put the data, and 1 for Freq:. Arrow down to [latex]\mu[/latex]: and arrow over to [latex]> \mu 0[/latex]. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value ([latex]p = 0.0396[/latex]) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation. [latex]\mu > 65[/latex] is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with [latex]t = 1.9781[/latex] (test statistic) and [latex]p = 0.0396[/latex] (p-value). Make sure when you use Draw that no other equations are highlighted in [latex]Y =[/latex] and the plots are turned off.

Hypothesis Tests for a Single Population Proportion

Still, with the hypothesis tests for a single population proportion, the by-hand method will still be shown as it did in 8.5 Additional Information and Full Hypothesis Test Examples. The same thing as before, doing hypothesis tests is a lot more than just plugging into the calculator - you need to know the process involved. You will see a box that separates the calculator information.

Example

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.

Solution

Set up the hypothesis test:

The 1% level of significance means that α = 0.01. This is a test of a single population proportion.

[latex]H_{0}: p = 0.50, H_{a}: p \neq 0.50[/latex]

The words "is the same or different from" tell you this is a two-tailed test.

Calculate the distribution needed:

Random variable: [latex]P^{\prime}[/latex] = the percentage 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.

[latex]{P}^{\prime } \sim N\left(p,\sqrt{\frac{p\cdot q}{n}}\right)[/latex] Therefore, [latex]{P}^{\prime } \sim N\left(0.5,\sqrt{\frac{0.5\cdot 0.5}{100}}\right)[/latex]

where [latex]p = 0.50[/latex], [latex]q = 1−p = 0.50[/latex], and [latex]n = 100[/latex].

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

[latex]\text{p-value} = 0.5485[/latex]

where [latex]x = 53[/latex], [latex]p^{\prime} = \frac{x}{n} = \frac{\text{53}}{100} = 0.53[/latex].

Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion [latex]p^{\prime}[/latex] is 0.53 or more OR 0.47 or less (see the graph below).

A normal distribution with 0.50 in the middle and values of 0.47 and 0.53 with shaded regions in both tails. 1/2(p-value) = 0.27425 — **Figure 5.** 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. 1/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.

[latex]\mu = p = 0.50[/latex] comes from [latex]H_{0}[/latex], 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 [latex]\alpha[/latex] and the p-value:

Since [latex]\alpha = 0.01[/latex] and [latex]\text{p-value} = 0.5485[/latex]. [latex]\alpha \lt \text{p-value}[/latex]

Make a decision: Since [latex]\alpha \lt \text{p-value}[/latex], you cannot reject [latex]H_{0}[/latex].

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 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.)

Using the TI-83, 83+, 84, 84+ Calculator

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 ([latex]p = 0.5485[/latex]) 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 [latex]z = 0.6[/latex] (test statistic) and [latex]p = 0.5485[/latex] (p-value). Make sure when you use Draw that no other equations are highlighted in [latex]Y =[/latex] and the plots are turned off.

The following examples show some helpful screenshots from the TI-83 or TI-84 graphing calculator.

Example

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; 0.98; 0.98 1.02; 0.95; 0.95

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.

Solution

Let’s follow a four-step process to answer this statistical question.

State the Question: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be [latex]H_{0}: \mu \le 1, H_{a}: μ > 1[/latex]
Plan: We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
Do the calculations: We will input the sample data into the TI-83 as follows.
State the Conclusions: Since the p-value* (p = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

TI calculator screen with values in L1: 1.08, .98, .98, 1.02, .95, .95 — **Figure 6.** List feature being used for the hypothesis test

TI calculator screen of T-Test with Data highlighted, mu:1, and >mu0. — **Figure 7.** T-Test showing Data being used with values filled in to run the hypothesis test.

TI calculator screen of T-Test with values. — **Figure 8.** T-Test screen with alternate hypothesis, test statistic, p-value, point estimate, standard deviation, and n.

TI calculator screen showing a normal distribution with a test statistics and p-value. — **Figure 9.** Normal Distribution curve with test statistic, z=2.0138, and p-value of approximately 0.0359

Example

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.

Solution

We will follow the four-step process.

We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be [latex]H_{0}: p \le 0.00034[/latex]and [latex]H_{a}: p > 0.00034[/latex].
If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.
We will be testing a sample proportion with [latex]x = 172[/latex] and [latex]n = 420,019[/latex]. The sample is sufficiently large because we have [latex]np = 420,019(0.00034) = 142.8[/latex], [latex]nq = 420,019(0.99966) = 419,876.2[/latex], two independent outcomes, and a fixed probability of success [latex]p = 0.00034[/latex]. Thus we will be able to generalize our results to the population.
The associated TI results are below.
Since the [latex]\text{p-value} = 0.0073[/latex] is greater than [latex]\alpha = 0.005[/latex], we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.

TI calculator screen of 1-PropZTest with values. — **Figure 10.** 1-PropZTest screen with alternate hypothesis, test statistic, p-value, point estimate, and n.

Your Turn!

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.

Solution

We will follow the four-step plan.

We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.
Since we are presented with proportions, we will use a one-proportion z-test. The hypotheses for the test will be [latex]H_{0}: p = 0.00078[/latex] and [latex]H_{a}: p \neq 0.00078[/latex].
The following screenshots display the summary statistics from the hypothesis test.
Since the p-value, [latex]p = 0.00063[/latex], is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.

Helpful Videos for the Graphing Calculator

Below are links to helpful videos for using the graphing calculator for the concepts covered on this page:

Additional Technology Tools

In addition to the graphing calculator, there are some additional technology tools that can be used for the concepts covered on this page. Below are links to helpful videos for those tools:

Microsoft Excel:

Section Practice

Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal.

Is this a test of means or proportions?
What symbol represents the random variable for this test?
In words, define the random variable for this test.
Is the population standard deviation known and, if so, what is it?
Calculate the following:
- [latex]\overline{x} = \underline{\hspace{2cm}}[/latex]
- [latex]\sigma = \underline{\hspace{2cm}}[/latex]
- [latex]s_{x} = \underline{\hspace{2cm}}[/latex]
- [latex]n = \underline{\hspace{2cm}}[/latex]
Since both [latex]\sigma[/latex] and [latex]{s}_{x}[/latex] are given, which should be used? In one to two complete sentences, explain why.
State the distribution to use for the hypothesis test.

A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time on death row could likely be 15 years.

Is this a test of one mean or proportion?
State the null and alternative hypotheses.
Is this a right-tailed, left-tailed, or two-tailed test?
What symbol represents the random variable for this test?
In words, define the random variable for this test.
Is the population standard deviation known and, if so, what is it?
Calculate the following:
- [latex]\overline{x} = \underline{\hspace{2cm}}[/latex]
- [latex]s = \underline{\hspace{2cm}}[/latex]
- [latex]n = \underline{\hspace{2cm}}[/latex]
Which test should be used?
State the distribution to use for the hypothesis test.
Find the p-value.
At a pre-conceived [latex]\alpha = 0.05[/latex], what is your:
- Decision:
- Reason for the decision:
- Conclusion (write out in a complete sentence):

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

Is this a test of one mean or proportion?
State the null and alternative hypotheses.
Is this a right-tailed, left-tailed, or two-tailed test?
What symbol represents the random variable for this test?
In words, define the random variable for this test.
Calculate the following:
- [latex]x = \underline{\hspace{2cm}}[/latex]
- [latex]n = \underline{\hspace{2cm}}[/latex]
- [latex]{p}^{\prime } = \underline{\hspace{2cm}}[/latex]
Calculate [latex]\sigma_{x} = \underline{\hspace{2cm}}[/latex]. Show the formula set-up.
State the distribution to use for the hypothesis test.
Find the p-value.
At a pre-conceived [latex]\alpha = 0.05[/latex], what is your:
- Decision:
- Reason for the decision:
- Conclusion (write out in a complete sentence):

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?