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Chapter 8 Hypothesis Test

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Steps to a Hypothesis Test 1.Hypotheses –Null Hypothesis (Ho) –Alternative Hypothesis (Ha) 2.Alpha 3.Distribution (aka model) 4.Test Statistics and P-value 5.Decision 6.Conclusion

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Steps to a Hypothesis Test Can remember the steps by the sentence: “Happy Aunts Make The Darndest Cookies”

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Example 1– Hypothesis Testing An attorney claims that more than 25% of all lawyers advertise. A sample of 200 lawyers in a certain city showed that 63 had used some form of advertising. At α = 0.05, is there enough evidence to support the attorney’s claim?

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Hypotheses (Sets up the two sides of the test) 1.Build the Alternative Hypothesis (Ha) first. –based on the claim you are testing (you get this from the words in the problem) Three choices –Ha: parameter ≠ hypothesized value –Ha: parameter < hypothesized value –Ha: parameter > hypothesized value 2.Build Null Hypothesis (Ho) next. –opposite of the Ha (i.e. =, ≥, ≤ )

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Example 1– Constructing Hypotheses We need to know what parameter we are testing and which of the three choices for alternative hypothesis we are going to use. –“An attorney claims that more than 25% of all lawyers advertise” tells us that this is a test for proportions so our parameter is p. –“claims that more than 25%” tells us that Ha: p >.25 and therefore Ho: p ≤.25

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Alpha Alpha = α = significance level –How much proof we are requiring in order to reject the null hypothesis. –The complement of the confidence level that we learned in the last chapter –Usually given to you in the problem, if not, you can choose. Most popular alphas: 0.05, 0.01, and 0.10

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Example 1 – Alpha “At α = 0.05” is given to us in the problem so we just copy α = 0.05

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Model The model is the distribution used for the parameter that you are testing. These are just the same as we used in the confidence intervals. –p and μ (n ≥ 30) use the normal distribution –μ (n < 30) uses the t-distribution – uses the chi-squared distribution

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Example 1 - Model The model used for a proportion is the normal.

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Test Statistic You will have a different test statistic for each of the four different parameters that we have learned about. –p : – μ (n ≥ 30) :

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Test Statistic You will have a different test statistic for each of the four different parameters that we have learned about. –μ (n < 30) : – :

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p-value This is the evidence (probability) that you will get off of your chart and then compare against your criteria (alpha). You will need to find the appropriate probability that goes with your Ha. –> and < Ha’s are called one-tailed tests. –≠ Ha’s are called two-tailed tests. For z and χ 2 you have to take the > probability X2

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Example 1 – Test Statistic and p-value The formula for a test statistic for proportions is: So, from our problem we need a proportion from a sample (p-hat), the proportion from our hypothesis (p o ), and a sample size (n).

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Example 1 – Test Statistic and p-value “A sample of 200 lawyers in a certain city showed that 63 had used some form of advertising” tells us that –p-hat = 63/200 or 0.315 From our hypothesis we know –p o = 0.25 (which means that q o = 0.75) “sample of 200” tells us that –n = 200

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Example 1 – Test Statistic and p-value So our test statistic and p-value are

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Decision – (always about Ho) We have two choices for decision –Reject Ho –Do Not Reject Ho If our evidence (p-value) is less than α we REJECT Ho. If our evidence (p-value) is greater than α we DO NOT REJECT Ho.

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Example 1 - Decision Our p-value is 0.0170 and our alpha is 0.05 –So, since our p-value is less than our alpha our decision is: REJECT Ho.

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Conclusion – (always in terms of Ha) Conclusions –Reject Ho “There is enough evidence to suggest (Ha).” –Do Not Reject “There is not enough evidence to suggest (Ha).”

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Example 1 - Conclusion Our decision to was to reject Ho, so our conclusion is: “There is enough evidence to suggest that p>0.25”

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Example 1 - Summary 1.Ho: p ≤ 0.25 Ha: p > 0.25 2.α = 0.05 3.Model: Normal 4.z = 2.12 and p-value = 0.0170 5.Reject Ho 6.There is enough evidence to suggest that p>0.25.

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Example 2 – Hypothesis Testing A researcher reports that the average salary of assistant professors is more than $42,000. A sample of 30 assistant professors has a mean of $43,260. At α = 0.05, test the claim that assistant professors earn more than $42,000 a year. The standard deviation of the population is $5230.

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Example 2 (cont.) Hypotheses –Ho: μ ≤ $42,000 –Ha: μ > $42,000 (given claim is “more than”) Alpha –α = 0.05 (given) Model –Normal (n ≥ 30 and it’s a mean)

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Example 2 (cont.) Test statistic and p-value:

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Example 2 (cont.) Decision –0.0934 > 0.05 (p-value > alpha) –DO NOT REJECT Ho Conclusion –We do not have evidence to suggest that μ > $42,000.

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Example 3 – Hypothesis Testing A physician claims that joggers’ maximal volume oxygen uptake is greater than the average of all adults. A sample of 15 joggers has a mean of 40.6 milliliters per kilogram (ml/kg) and a standard deviation of 6 ml/kg. If the average of all adults is 36.7 ml/kg, is there enough evidence to support the physicians claim at α = 0.05?

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Example 3 (cont.) Hypotheses –Ho: μ ≤ 36.7 –Ha: μ > 36.7 Alpha –α = 0.05 (given) Model –t (14)

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Example 3 (cont.) Test statistic and p-value:

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Example 3 (cont.) Decision –(0.01,0.025) < 0.05 (p-value < alpha) –REJECT Ho Conclusion –There is evidence to suggest that μ > 36.7.

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Example 4 – Hypothesis Testing A researcher knows from past studies that the standard deviation of the time it takes to inspect a car is 16.8 minutes. A sample of 24 cars is selected and inspected. The standard deviation was 12.5 minutes. At α=0.05, can it be concluded that the standard deviation has changed?

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Example 4 (cont.) Hypotheses –Ho: σ = 16.8 –Ha: σ ≠ 16.8 Alpha –α = 0.05 (given) Model –χ 2 (23)

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Example 4 (cont.) Test statistic and p-value:

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Example 4 (cont.) Decision –(0.05,0.10) > 0.05 (p-value > alpha) –DO NOT REJECT Ho Conclusion –There is not enough evidence to suggest that σ ≠ 16.8.

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