## 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.