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Hypothesis Testing. Tests of Significance. 10.1 confidence interval to estimate population parameter 10.2 Test some claim about the population based on sample data. An Example.
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Tests of Significance • 10.1 confidence interval to estimate population parameter • 10.2 Test some claim about the population based on sample data.
An Example • Suppose a company claims the mean contents of a cereal to be 16 oz. You believe the mean contents to be less than 16 oz. Suppose you buy a random sample of 49 boxes, and find their mean contents to be 15.90 oz. Assuming we know . Comment on the company’s claim.
Solution • We assume the company’s claim is true and see how likely our sample is… • Does our sample mean of 15.90 oz. reflect a true lack of contents ? or • Could we have gotten a mean of 15.90 oz. just by chance?
Solution • We start with the null hypothesis • Original claim made about the population. • State the alternative to the null hypothesis We say this test is one-sided since the alternate is on one side of the mean.
Solution • Compute Test Statistic. In this case since we are investigating a population mean( ) with a known standard deviation( ) we use a Z-Test
Solution • Chance we got a sample with at most 15.90 oz is very small. • This provides evidence we should reject the null hypothesis claim that the cereal contents contain 16 oz. • There is evidence that the true mean is less than 16 oz.
More on P-values Left tail test Right tail test • P-Value • Probability of observing any sample with a value as extreme or more extreme than the one from the sample data(in the problem). • Smaller P-value More evidence to reject the null hypothesis Two-sided test
Significance Level • Saying the p-value should be small to reject null hypothesis needs to be more defined. • Each problem will provide you with an area that the p-value must be less than in order to prove your sample is significant evidence against the null hypothesis
Right Tail Test • Suppose 150 college students were asked their Intelligence Quotient (IQ). From the data you find out that and assume we know A college professor says that he knows the population mean is 116. Using a 0.05 significance level, determine if the professor may be correct.
Right Tail Test Continued • Stating the null and alternative hypothesis We assume the professor is right. We will check to see if our sample provides evidence against the null hypothesis • Notice the null and alternative hypothesis always look at population parameters. “Right tail test” since we the alternate hypothesis is > than the null hypothesis
Right Tail Test Continued • Compute the z-test statistic • Since we are looking a population mean with a known standard deviation…
Right Tail Test Continued • Calculate the p-value • Since this is a right tail test… • Interpret p-value at significance level • Since .0006 < 0.05 we reject the null hypothesis. The professor is wrong.
Two-sided Test • At times we don’t care if the alternative hypothesis is > or < to the population parameter • In this case we conduct a two-sided test.
Two-Sided Example • Logan is doing a project for his local city planning department in which he tests the claim the city’s housing project contains family units of average size. A SRS of 80 homes in the project shows a sample mean of 3.4 people per unit. Assume the standard deviation is 1.4 people. Construct a hypothesis test to determine whether the average size of family units in the housing project is different from the national average of 3.8. Use and the 4 steps
Solution • The population is all the families in the city. We will investigate the average family size(unit) in the city. We will conduct a hypothesis with the following: The mean of the city is equal to the national average of 3.8 The mean of the city differs from the national average of 3.8
Solution • We will conduct a two-sided z-test since we know standard deviation of the population. We will check the conditions below: (1) Data is from an SRS (2) The sampling distribution of will be normal by central limit theorem since n > 30.
Solution The p-value associated with this z-score is: 2(.0052) = .0104
Solution • Our p-value is .0104. This is greater than the significance level . We fail to reject the null hypothesis at the 1% significance level. We do not have evidence at this significance level that the city’s mean differs from the national average.
Rejection Regions • There is another way to decided whether we reject or accept the null hypothesis • We will calculate the z-score for the significance level and compare that to the z-statistic from our sample.
Rejection Region right tail Suppose we have This is the z-stat from yesterday
2-sided Rejection Region • Suppose