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Hypothesis Testing When “p” is small, we reject the Ho.

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Warm-up: state your hypotheses a) Larry's car averages 26 miles per gallon on the highway. He switches to a new brand of motor oil that is advertised to increase gas mileage. After driving 3000 highway miles with the new oil, he wants to determine if the average gas mileage has increased. = the mean gas mileage for Larry ’ s car on the highway. Ho: = 26 mpg Ha: > 26 mpg

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(b) A May 2005 Gallup Poll report on a national survey of 1028 teenagers revealed that 72% of teens said they rarely or never argue with their friends. You wonder whether this national result would be true in your school. So you conduct your own survey of a random sample of students at your school. Warm-up: state your hypotheses == the proportion of teens in your school who rarely or never fight with their friends. Ho: ==.72 Ha: ≠≠.72

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The city manager noted that paramedics arrived within 8 minutes after 78% of all calls involving life-threatening injury last year. Based on this year's random sample of 400 calls, she wants to determine whether the paramedics have arrived within 8 minutes more frequently this year. p = proportion of calls involving life–threatening injuries for which the paramedics arrived within 8 minutes. H 0 : p = 0.78 H a : p > 0.78

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A national study reports that households spend an average of 30% of their food expenditures in restaurants. A restaurant association in your area wonders if the national results apply locally. They interview a sample of households and ask about their total food budget and the amount spent in restaurants. μ = mean percent of local household food expenditures used for restaurant meals. H 0 : μ = 30 H a : μ ≠ 30

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Why is “p” always gets to decide which hypothesis needs to go Small P-values are evidence against H 0 because they say that the observed result is unlikely to occur when H 0 is true. Large P-values fail to give evidence against H 0. = the mean gas mileage for Larry ’ s car on the highway. Ho: = 26 mpg Ha: > 26 mpg

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Two-tailed test One study chose 18 subjects at random from a group of people who assembled electronic devices. Half of the subjects were assigned at random to each of two groups. Both groups did similar assembly work, but one work setup allowed workers to pace themselves, and the other featured an assembly line that moved at fixed time intervals so that the workers were paced by machine. After two weeks, all subjects took the Job Diagnosis Survey (JDS), a test of job satisfaction. Then they switched work setups and took the JDS again after two more weeks. This is a matched pairs design. The response variable is the difference in JDS scores, self-paced minus machine- paced. Suppose we know that differences in job satisfaction scores in follow a Normal distribution with standard deviation σ = 60, perform a hypothesis testing for this case.

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Part1: Hypotheses Ho: µ = 0, Ho: µ = 0, there is NO difference in job satisfaction between the two work environments Ho: µ ≠ 0, Ho: µ ≠ 0, there is a difference in job satisfaction between the two work environments

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Part 2: Conditions Because the workers were chosen without replacement, randomly sampled from the assembly workers, and then randomly assigned to each group, the requirements of SRS and independence are met. The question states that the differences in job satisfaction follow a Normal distribution.

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Part 3: Test Statistic x = 17 µ = 0 ∂ = 60 n = 18 z= x- ℳ σ/√n z= 17-0 60/√18 z= 1.20The two-sided p-value for z=1.20 is p = 2(.1151) p =.2302

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Part 4: Conclusion With a p-value of.2302, we have a fairly strong evidence against the alternative hypothesis therefore we can conclude that there is no significant job difference satisfaction between the two work environments set by the company. Ho: µ = 0, Ho: µ = 0, there is NO difference in job satisfaction between the two work environments Ho: µ ≠ 0, Ho: µ ≠ 0, there is a difference in job satisfaction between the two work environments REJECT p =.2302

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Your turn! Job satisfaction with a larger sample: Suppose that the job satisfaction study had produced exactly the same outcome x = 17 as in the previous example but from a sample of 75 workers rather than just 18 workers. perform a hypothesis testing for this new case. Z = 2.45 P =.0142 we can conclude that there is a significant job difference satisfaction between the two work environments set by the company. A p=value of.0142 gives a strong evidence against the null hypothesis, therefore we can conclude that there is a significant job difference satisfaction between the two work environments set by the company.

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The company medical director institutes a health promotion campaign to encourage employees to exercise more and eat a healthier diet. One measure of the effectiveness of such a program is a drop in blood pressure. The director chooses a random sample of 50 employees and compares their blood pressures from physical exams given before the campaign and again a year later. The mean change in systolic blood pressure for these n = 50 employees is x = −6. We take the population standard deviation to be σ = 20. The director decides to use an α = 0.05 significance level.

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Step 1: Hypotheses We want to know if the health campaign reduced blood pressure on average in the population of all employees at this large company. Taking μ to be the mean change in blood pressure for all employees, we test H 0 : μ = 0 H a : μ < 0

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Step 2: Conditions Since σ is known, we will use a one- sample z test for a population mean. Now we check conditions. SRS The medical director took a “ random sample ” of 50 company employees. Our calculation method assumes that an SRS was taken. Normality The large sample size (n = 50) guarantees approximate Normality of the sampling distribution of x, even if the population distribution of change in blood pressure isn't Normal. Independence There must be at least (10)(50) = 500 employees in this large company since the medical director is sampling without replacement.

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Step 3: Test Statistic P = P(Z ≤ −2.12) = 0.0170

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Step 4: Interpretation Since our P-value, 0.0170, is less than α = 0.05, this result is statistically significant. We reject H 0 and conclude that the mean difference in blood pressure readings from before and after the campaign among this company's employees is negative. In other words, the data suggest that employees' blood pressure readings have decreased on average. H 0 : μ = 0 H a : μ < 0 α = 0.05 P = 0.0170 μ is the mean change in blood pressure for all employees REJECT

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