# Practical Statistics Z-Tests. There are six statistics that will answer 90% of all questions! 1. Descriptive 2. Chi-square 3. Z-tests 4. Comparison of.

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Practical Statistics Z-Tests

There are six statistics that will answer 90% of all questions! 1. Descriptive 2. Chi-square 3. Z-tests 4. Comparison of Means 5. Correlation 6. Regression

Z-test are for proportions. This test is so easy…. That it is not even given in some stat programs like SPSS…..

Z-test are for proportions. What is the probability that out of 25 customers, 22 would like the service when the usual percent is that 70% are satisfied?

Z-test are for proportions. They come in two types: 1. A sample proportion against a hypothesis.

Z-test are for proportions. They come in two types: 1. A sample proportion against a hypothesis. 2. Two samples compared to each other.

Z-test are for proportions. The standard error for proportions is: Where p = freq/total and q = 1 - p

Z-test are for proportions. Hence: Where p is the hypothesized value, and p t is the proportion found in a sample of size n.

Z-test are for proportions. Suppose that XYZ Company believed that 20% of their customers bought 80% of their product. A sample of 200 customers found that 25% bought 80% of the product. Was the company correct in their estimate?

Z-test are for proportions. The test statistic looks like this:

Z-test are for proportions. Since the test was “two-tailed,” the critical value of Z would be 1.96. Therefore, we would conclude that there is not enough evidence to over-ride the assumption that 20% of the customers bought 80% of the product.

Z-test are for proportions. http://faculty.vassar.edu/lowry/tabs.html#z P = 0.077

Z-test are for proportions. They come in two types: 1. A sample proportion against a hypothesis. 2. Two samples compared to each other.

Z-test are for proportions. The test for this case looks like this:

Z-test are for proportions. Usually, the test assumes that the two groups Are equal, or:

There is a problem here. What is the value of: ?

p is the value of the population proportion, but we usually don’t know that value, so p is estimated by the weighted average of the two groups….

Suppose that a new product was test marketed in the United States and in Japan. The company hypothesizes that both countries response to the product will be the same. 80% of a sample of 500 said they would buy the product again in the U.S., while 75% of a sample of 200 in Japan said they would buy the product again.

Test the hypothesize…..

The test would be: But what is p?

Since p = 0.80 in the U.S., and 0.75 in Japan, the weighted average is used for p. So: P = ((.8 x 500)+(.75x200))/700 = 0.786

The test would be: Z =.05/.0343 = 1.45 The critical value is 1.96; The probability of z = 1.45 is P = 0.147. The U.S. and Japanese customers are the same.

Questions: 1.Jason works in a building with four floors served by an elevator. He has to use the elevator several times a day to carry heavy material from one floor to the next. He states, only partly in jest, that the elevator hates him because it is never on the floor he is on, and he always has to wait for it to come. To test his hypothesis, he kept track of where the elevator was for one month when he used it. He had 60 recorded elevator uses and 9 times the elevator was at his floor when he needed it. Test Jason’s hypothesis. 2. Suppose Jason always needs to start at the second floor so, along with the info above, he also kept track of which floor the elevator was actually on when he needed it. He found: first floor: 20; second floor: 9; third floor, 12; and fourth floor: 19. Would the answer to #1 be changed if Jason worked on a different floor?

Questions: 1. Women buy 20% more BicTac than men. A new company produces a new version of BicTac supposedly targeted at men. A MR survey of 200 users found that 100 customers of the new product were women. Has the new product made any inroad into the male market? 2. An administrator believes that the number of men and women in an MBA program is the same, in a sample of 500, how many more women than men must be found to reject the administrator’s hypothesis?

Service Encounter and Personality Recall that 60% of shoppers normally are women, but this sample found that 55% were women. Is the sample bad? Z = (.55 -.60)/SQ((.6 X.4)/271) = -1.68 p = 0.093

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