FINDING BETA (b)

~A very tricky problem, so pay close attention to the procedure

~Let's review the types of errors:

 TYPE I : Rejecting a true null hypothesis
 TYPE II: Accepting a false null hypothesis
                (or rejecting a true alternate hypothesis)

~ALPHA (a) = The probability of a type I error = significance level
                   (usually .05 or .01). (area of the rejection region)

~BETA (b) = The probability of a type II error (not easy to find)

~Note: Ideally, we would like both types of errors as small as possible, however,  the smaller we choose alpha, the larger the beta. So, we need to decide what's more important for us in our study.  Usually, alpha is considered more important.

~Note: If the claim in the alternate hypothesis is relatively close to the one in the null hypothesis, the chances of making a type II error increases. (rejecting the alternate claim when it is true). On the other hand, the further away the alternate claim is from the null hypothesis, the smaller the beta.

~Note: When a normal curve is drawn with a given mean and standard deviation, we assume that it is the true mean.

~Note: If we draw another normal curve with a different mean, then we are assuming that this new mean is the true mean. This is important to understand in order to find beta.

~So, how does one find beta for a given alpha?

~Let's assume a one-tail test for our discussion with Ho: m=m_o, H₁: m < m_o.

~Also, let's assume that the population standard deviation is known (test statistic will be a z-value). If not, then the test statistic will be a t-value.
(see link "What test to Use")

~Let's assume that the motivation for choosing H₁ lies in the fact that our sample gave a mean less than m_o claimed, call it m₁.

~Now, set up a normal curve with the original m_o as the mean (we are assuming this to be the true mean).  Using a given alpha (.05 or .01), get the critical value of z= -1.645 for alpha=.05 or z= -2.33 for alpha = .01. These are fixed & depend on the significance level only.

~Note: Critical t-values will not be equal to the above for the same significance levels since there are many different t-curves (one for each sample size n). To get a specific t-curve, subtract 1 from n for the degrees of freedom defining that curve. For example, for n=20, the df=19 & we read that row to get the critical values (t=-1.73 & t=-2.54 for the .05 & .01 levels, respectively). Also, note that the last row in the table gives values consistent with the z-values from the standard normal curve since t-curves approach the standard normal curve as n increases.

~Using the critical z-value, convert it an actual score (use the formula relating the two).  Call it m₂. Therefore, since this is a left tailed test, we will be accepting H_o for all scores to the right of m₂.

~Using the sample mean, m₁, above, set up a normal curve assuming this value as the true mean (we are now assuming this mean to be the true mean) and locate the other score m₂.

~Beta will be the area (probability) to the right of the score m₂. That way we are accepting m_o (when m₁ is assumed true).

~Example: The Manufacturer of a patent medicine claimed that it was 90% effective in relieving an allergy for a period of 8 hours. In a random sample of 200 people who had the allergy, the medicine provided relief for 160 people (80%). Determine whether the manufacturer's claim is legitimate at the 0.01 level of significance & also determine beta (probability of a type II error).

~Use your TI-83: STAT, TESTS, 1-PropZTest(menu 5), Enter, Po: 0.9 (null hypothesis), x: 160, n: 200, choose < Po for the alternate hypothesis (our sample suggests that), Calculate, Enter.

Look for the P-value .0000012 (written in scientific notation). Since this P-value is much less than the significance level (0.01), we reject the null hypothesis (manufacturer's claim).

~Now for beta. If we set up a standard normal curve with 0.9 assumed to be true (at Z=0) & use a left-tailed test at the 0.01 level of signifcance & find the critical value (Z score at the right of the rejection region & to the left of 0.9), we would get Z = -2.333. Converting this Z score to an actual percentage score, we would get .85 (Use -2.333=(x-.9)/sqr[(.9)(.1)/200], then solve for x). This means, we woud reject the null hypothesis of .90 if our test statistic (from our sample) is less than .85. In this case, it was .80. That's why we rejected the null hypothesis.

~Let's assume, in reality, that the true proportion is 0.80 (from our sample) instead of 0.90. Now, set up another standard normal curve with .80 assumed true (at Z=0). Then locate .85 to the right of it. We need to convert .85 to a Z score (see note below) (use Z=(.85-.80)/sqr[(.8)(.2)/200]) under the assumption that .80 is the true proportion.

That will gives us, Z = 1.77. Now, find the area to the right of this score, uses, 2nd, VARS, menu 2, insert (1.77, 10000). That way, we are rejecting the true proportion of .80 (2nd standard normal sketch) but, since we are finding the area to the right of .85, we are not rejecting the claim of .90 (1st standard normal sketch). This is beta (.038).

~Note: You can by pass converting .85 to a Z score by doing the following:
2nd VARS, menu 2, insert (.85,10000,.8,sqr[(.8)(.2)/200]).

~Note: The POWER of a test is defined as the probability of rejecting a false null hypothesis (making the correct decision). So, beta would be 1 - the POWER of a test, since this would represent the probability of "accepting a false Ho". Since rejecting the null hypothesis depends on the location of the test statistic and, in turn, the test statistic depends on the sample size n (see formulas for the test statistic), increasing the sample size n, increases the POWER of a test, thus decreasing beta.

~I will not hold you responsible for this type of problem, however, it serves as a very good problem for extra credit.