6.2 Tests of Significance

FormalizingWe saw in the last section how to find a confidence interval. In this section, we use the confidence interval to come up with a formal test to be able to say whether or not we think a sample is representative of the population. We assume n is at least 30, and hence may use s to estimate σ. First some terms.The null hypothesis is a statement asserting no change, no difference, or no effect. The null hypothesis is labeled H0.In other words, the null hypothesis can be thought of as what we “normally” expect.

The alternative hypothesis is a statement that might be true instead of the null hypothesis. It usually contains the symbol >, <, or ≠. The alternate hypothesis is labeled Ha.Let’s think back to our example from 5.2 about a teacher who used a new biology book and wanted to see if it improved the scores of her students. We can use the language above, as well as some new language, to formalize our conclusion on the board.

We should choose a level of significance since it is unreasonable to assume that any sample should have exactly the same mean.This level of significance is subject to the data, what is being tested, and the desires or needs of the researchers. A value of .05 is often selected. We will use that. Let us continue.

Hypothesis Testing

The process we have discussed above is now summarized for Hypothesis Testing:

1. Identify null and alternate hypothesis.2. Choose the level of significance.3. Select the test statistic and determine its

value.4. Determine the critical region.5. Make your decision.

ExamplePrior to expanding the faculties of a town library, it was determined that 385 books per day were loaned out. It was believed that a new addition to the building would increase the mean number of books loaned out per day. After completion of the new addition, a random sample of 35 days showed a mean of 395 and a standard deviation of 26 books loaned out per day. At a 10% level of significance, is there sufficient evidence to indicate that more books are being loaned out per day on average?

Practice Problem

Sonnets by a certain Elizabethan poet are known to contain an average of μ=8.9 new words (words not used in the poet’s other works) with standard deviation σ=2.5. Now a manuscript with 46 new sonnets has come to light , and scholars are debating whether it is the poet’s work. The new sonnets contain an average of 9.2 words not used in the poet’s known works. What can we conclude about the authorship of the new poems? Use α=.02.

z-TestsThere are three general forms that the alternate hypothesis Ha can take

1. Ha: μ≠a 2. Ha: μ>a3. Ha: μ<a• These are respectively called a two-tailed test, a right-

tailed test, and a left-tailed test. We can see why by drawing the pictures on the board.

• All these tests use the standard normal variable z, and we refer to such a test as a z-test.

Two-tailed Test

A machine usually packs 500 nails on average into each box, and we would like to test this claim. Suppose a random sample of 39 boxes produces a mean of 498.7 nails per box with a standard deviation of 14.1 nails. Is there reason to believe that the mean is different than 500? Use α=.01.

Now all a z-test cares about is if you fell into the critical region or not. You could be very close to a critical value or 17 standard deviations away from it and you would still obtain the same result using a z-test.P-values, in contrast, attempt to remedy this problem.

The P-value is the probability of obtaining a z-value greater than or equal to the one which was obtained. For example, if z=2.12, then our P-value is P(z≥2.12)=.5-.4830=.017. Thus the smaller the P-value, the stronger the evidence in support of the alternate hypothesis.

The level of calcium in the blood in healthy young adults varies with mean about 9.5 milligrams per deciliter and standard deviation .4. A clinic in rural Guatemala measures the blood calcium level of 160 healthy pregnant women at their first visit for prenatal care. The mean is 9.57. Give the P-value and a 95% confidence interval.

How small is small?

Now we run into the problem of determining what we mean by a “small enough” P-value. One option that is usually put forth is to compare our P-value with our level of significance α. The P-value approach to testing hypotheses: When testing hypotheses at a level of significance α, we reject the null hypothesis if P≤ α. We do not reject H0 if P>α.If the P-value is as small or smaller than the specified value α, then we say that the data is statistically significant at significance level α.

A tire gauge is designed to measure 30 pounds square inch (psi). The manufacturer was concerned that the gauges were not meeting specifications. A sample of 50 readings gave a sample mean 30.1 psi and a standard deviation of .35. Find the P-value and determine which of the following levels of significance H0 would be rejected: α=.1, α=.05, α=.01.

Practice Problem

Sonnets by a certain Elizabethan poet are known to contain an average of μ=8.9 new words (words not used in the poet’s other works) with standard deviation σ=2.5. Now a manuscript with 46 new sonnets has come to light , and scholars are debating whether it is the poet’s work. The new sonnets contain an average of 9.2 words not used in the poet’s known works. Find the p-value.