## 1. Intro

In this sub-competency we learn how to carry out a hypothesis test for a parameter from just one population. We will learn two techniques, one for investigating a population proportion (p) and one for investigating a population mean (μ). Recall from sub-competency 9 that there are four main steps in the process:

1. State the null and alternative hypotheses.
2. Calculate the test statistic, which tells us how “different” our sample results are from what is being claimed in the null hypotheses.
3. Determine the probability and significance level of our sample producing the results that it did. In other words, determine the probability, or likelihood, that a population with a parameter equal to the value stated in the null hypothesis would include a sample with the characteristic(s) of our sample.
4. Make and write the appropriate conclusions based upon the level of significance α for the hypothesis test.

The main computations will occur in Steps 2 and 3 above. If we are testing a population proportion, we’ll use the basic z-score as our test statistic, as it compares a sample result with the (claimed) population value relative to the standard deviation of the sample. To spoil all the fun, our test statistic is: A major difference occurs when we study a population mean. Most of you will need to return to sub-competency 8 (confidence intervals) where the t-distribution was introduced. Although population proportions follow standard normal distributions, population means, on the other hand, follow a somewhat wider distribution called the t-distribution. Therefore, our test statistic will be: Both of these values tell us how different, in terms of the standard deviation, our sample statistic is from the claimed value of the population parameter.

In sub-competency 8 we first created confidence intervals about a population proportion and then we modified the formulas slightly to create a confidence interval about a population mean. We follow the same path with hypothesis tests.

## Hypothesis Test for a Population Proportion

Our main goal is in finding the probability of a difference between a sample mean  and the claimed value of the population proportion, p0.

In order for the sampling distribution of a sample proportion  to be approximately normal with mean μ = and standard deviation the following 3 conditions need to be met:

• The sample was obtained through a simple random sample process.
• np ⋅ (1 – p) ≥ 10
• n ≤ 0.05 ⋅ N, where n is the sample size and N is the size of the population.

Example

In 1995, 40% of adults aged 18 years or older reported that they had “a great deal” of confidence in the public schools. On June 1, 2005, the Gallup Organization (www.gallup.com) released results of a poll in which 372 of 1004 adults aged 18 years or older stated that they had “a great deal” of confidence in public schools. Does the evidence suggest at the α = 0.05 significance level that the proportion of adults aged 18 years or older having “a great deal” of confidence in the public schools is significantly lower in 2005 than the 1995 proportion?

Step 1: State the null and alternative hypotheses.

Basically, the goal of this problem is to see whether attitudes about public schooling have changed over time. We are asked to use the results from 1995 as the “baseline” and see whether, ten years later, attitudes are lower. Thus:

• H0: p = 0.40
• Hα: p < 0.40

Notice that this is a one-tail test since the question in the example wants to know whether confidence levels are LOWER.

Step 2: Determine the level of significance.

We are asked to use α = 0.05.

Step 3: Calculate the test statistic.

We first need to identify the sample proportion and standard deviation from the information given in the problem. We see that: Using this information, the value of the test statistic is: So our sample proportion is just under 2 standard deviations below the claimed value of the population proportion.

Step 4: Determine the P-value and the level of significance.

Using a table of standard normal values with a z-value of z0 = -1.91 we find that the probability value is 0.0281. Using technology (which doesn’t do as much rounding as we do with our calculations), we find that the probability value is 0.0282691712.

The question provided us with a significance level of 5%. Thus, α = 0.05 .

Step 5: Make appropriate conclusions

Comparing our P-value with the level of significance, one can see that:

P-value = 0.0262 < α = 0.05.

Thus, we reject the null hypothesis, H0: p = 0.40. Our sample data provide significant evidence that the population proportion is not 0.40, and in fact, is likely much less. This means that significantly fewer people had “a great deal” of confidence in public schools in the year 2005 compared with the year 1995.

## 3. Using Confidence Intervals to Test Hypotheses

This is probably the easiest (and most understandable) method for testing a claim about a population proportion, as long as our null hypothesis contains the condition of not equal. Recall from sub-competency 8 that a confidence interval represents a range of values that we believe, with level C confidence, contains or captures the true population mean or proportion. We can use this interval to test the null hypothesis. The key to understanding this is to realize that a level C = (1 – α) ⋅ 100% confidence interval gives us the same results as a hypothesis test using a level of significance α. For example, a 95% confidence interval can be used in place of a hypothesis test using a significance level α = 0.05 = 5%.

To use a confidence interval, simply make the following observations:

• If our confidence interval contains the value claimed by the null hypothesis, then our sample result is close enough to the claimed value, and we therefore do not reject H0.
• If our confidence interval does not contain the value claimed by the null hypothesis, then our sample result is different enough from the claimed value, and we therefore reject H0.

A final note: The two main assumptions that we must have in order for the statistical calculations of hypothesis testing to be valid are (1) the sample data must be obtained through some random procedure, and (2) the sample data should (roughly) form a normal distribution with no strong skewness and no huge outliers. To check the (rough) normalcy of the data, you can simply create a stem-and-leaf plot and look at the overall pattern.

## 4. Testing Claims About the Population Mean

In this section we repeat the process of hypothesis testing, this time for when we are interested in testing claims made about a population mean, or average. We use the same process with only one slight modification: we must use a t-value for determining the difference between our sample mean and the (claimed) population mean. In a sense, we make the same modification that we made back in sub-competency 8, when we first learned about confidence intervals about a population proportion and then modified our work to fit population means.

Example

The US Food and Drug Administration recommends that individuals consume 1000 mg of calcium daily. The International Dairy Foods Association (IDFA) sponsors an advertising campaign aimed at male teenagers. After the campaign, the IDFA obtained a random sample of 50 male teenagers and found that the mean amount of calcium consumed was 1081 mg, with a standard deviation of 426 mg. Conduct a test to determine whether the campaign was effective. Use the α = 0.05 level of significance.

Step 1: State the null and alternative hypotheses.

The recommended level of calcium to consume is 1000 mg per day. This is what is expected (or believed) to occur. The IDFA sponsors a campaign aimed at male teenagers. Now, presumably, the goal of the IDFA is to increase the consumption of milk. (It doesn’t make sense for a dairy association to try to decrease milk consumption, does it?) Therefore, our hypotheses are:

• H0: p = 1000 mg
• Hα: p > 1000 mg

Note that we are conducting a one-tail hypothesis test. We are looking only for evidence that is significantly greater than what is being claimed.

Step 2: Calculate the test statistic.

The problem states that a sample of size n = 50 results in a sample mean calcium consumption level of 1081 mg and a sample standard deviation of s = 426 mg. Notice that these values are from our sample, and are not values from the population. This means we cannot use z-values. Our test statistic is: Therefore, our sample mean of 1081 mg is 1.34 standard deviations above the claimed population mean of 1000 mg. By now we should know that being 1.34 standard deviations different from the mean is not really that different (after all, roughly 95% of all data values lie within 2 standard deviations of the mean). So even before determining the P-value, we should have an idea that our data seem to provide evidence that the population mean may indeed be somewhere near 1000 mg, and, therefore, not significantly higher as the IDFA would have liked. In other words, we can guess our conclusion will be: fail to reject the null hypothesis H0. Step 3: Calculate the P-value and identify the level of significance.

To find the P-value, we use n – 1 degrees of freedom. Since our sample size is n = 50, our degrees of freedom is 49. There are two ways to find the P-values for the t-distribution: a table of significance values (shown here) or through technology. The problem with relying a table of values is that it’s not possible to have a list of significance values for every possible degrees of freedom value. For example, you’ll notice that a degrees of freedom value of 49 is not an option in the table shown. Therefore, to be on the safe side (in statistics, we always make conservative decisions), we go to the next lowest possible value for degrees of freedom, which happens to be 40. If we “rounded up” we’d be pretending that we actually have a larger sample size (and thus more information) than we actually do!

Moving along the row with 40 degrees of freedom, we see that our test statistic 1.34 falls between the two test statistics and their corresponding probabilities (located at the tops of the respective columns) as shown: Notice that the probabilities are “written” backward for standard math notation, since the larger probability (0.10) is on the left, and the smaller probability (0.05) is on the right. We simply rewrite this as:

0.05 < P-value < 0.10

Therefore, our P-value is larger than 0.05 and smaller than 0.10. Even though tables of significance values will rarely (if ever) provide us with an exact P-value, we can get a range of P-values that will lead us to making the “correct” conclusion.

Using technology we get a P-value of

P-value = 0.092489219

Notice that the far more approximate P-value does indeed fall inside the range of P-values we found from Table VI: 0.05 < 0.092… < 0.10.

For the level of significance α we look to the problem and see that α = 0.05.

Step 4: Make Appropriate Conclusions.

Using either the range of P-values from the table or the more precise P-value from technology, we observe that our P-value is larger than the level of significance α = 0.05. Therefore, we fail to reject the null hypothesis. Our conclusion would then be: “At the 5% significance level, our data do not provide enough evidence to conclude that the mean calcium consumption level has increased significantly due to the campaign from the IDFA.”

### Hypothesis Testing with the TI-83/84 Calculator:

All of these test can be found by hitting the [STAT] button and arrowing over to the TESTS menu.

Calculator Example 1: A population mean with statistics

A Sample of 38 cans is normally distributed with a mean weight of 12.5 oz and a population standard deviation of 2.8oz. At the 0.05 level of significance test the null hypothesis that the population mean is 14, that is Hο: μ=14, Hα: μ≠14 oz, with α=0.05.

Solution:

We choose [1:Z-Test] since we are using a z distribution. Enter the information as shown in screen 1 below, highlight [Calculate]and hit enter to get screen 2 or [Draw] to get screen 3. The p-value is 0.0082<α, so we reject the null hypothesis and write our conclusion.

Calculator Example 2: A population mean with data
A sample of 7 items is chosen from a normal distribution with the results{1,5,6,8,12,16,18}. Test the claim that µ<10 , that is Hο: μ=10, Hα: μ<10, with α=0.05.

Solution:

Here we are given the actual data from a sample we can have the calculator do all the work on the sample by entering the data into a list, say L1. We then choose [2: T-test]. Enter the information in screen 4 below, highlight [Calculate] and then hit ENTER to get screen 5 or choose [Draw] to get screen 6. The p-value of 0.4072> α, so we Fail to reject the null hypothesis and will need to write a conclusion that reflects this decision.

Calculator Example 3: A population proportion test
For a sample with 35 experiments 14 were successful. Test the claim that the experiment is successful more than 30% of the time, that is Hο:p = 0.3, Hα: p > 0.3, with α=0.05.

Solution:

We then choose [2: 1-PropZ-test]. Enter the information in screen 7 below, highlight [Calculate] and then hit ENTER to get screen 8 or choose [Draw] to get screen 9. The p-value of 0.0984> α, so we Fail to reject the null hypothesis and will need to write a conclusion that reflects this decision.

NOTE: x and n must be integers to use this test.