Ch. 9 examples

Summary of Hypothesis test steps Null hypothesis H0, alternative hypothesis H1, and preset α Test statistic and sampling distribution P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results 2-tailed ex H0: µ= 100 H1: µ ≠ 100 α = 0.05 Left tail ex H0: µ = 200 H1: µ < 200 Right tail ex H0: µ = 50 H1: µ > 50

Should you use a 2 tail, or a right, or left tail test? 2-tailed ex H0: µ= __ H1: µ ≠ __ Left tail ex H0: µ = ___ H1: µ < ___ Right tail ex H0: µ = __ H1: µ > __ Test whether the average in the bag of numbers is or isn’t 100. Test if a drug had any effect on heartrate. Test if a tutor helped the class do better on the next test. Test if a drug improved elevated cholesterol.

Type I and Type II error

Probabilities associated with error

Example #1- numbers in a bag Recall that I claimed that my bag of numbers had a mean µ = 100 and a standard deviation =21.9. Test this hypothesis if your sample size n= 20 and your sample mean x-bar was 90.

Ex #1- Hypothesis Test for numbers in a bag α = 0.05 Z = = P-value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #2– new sample mean for numbers in a bag If the sample mean is 95, redo the test: H0: µ = 100 H1: µ ≠ 100 α = 0.05 Z = = P-value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #3: Left tail test- cholesterol A group has a mean cholesterol of 220. The data is normally distributed with σ= 15 After a new drug is used, test the claim that it lowers cholesterol. Data: n=30, sample mean= 214.

Ex #3- cholesterol- test H0: µ 220 (fill in the correct hypotheses here) H1: µ 220 α = 0.05 Z = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #4- right tail- tutor Scores in a MATH117 class have been normally distributed, with a mean of 60 all semester. The teacher believes that a tutor would help. After a few weeks with the tutor, a sample of 35 students’ scores is taken. The sample mean is now 62. Assume a population standard deviation of 5. Has the tutor had a positive effect?

Ex #4: tutor Z = = P-value and/or critical value 4. Test conclusion H0: µ 60 (fill in the correct hypotheses here) H1: µ 60 α = 0.05 Z = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

9.2– t tests Just like with confidence intervals, if we do not know the population standard deviation, we substitute it with s (the sample standard deviation) and Run a t test instead of a z test

Ex #5– t test – placement scores The placement director states that the average placement score is 75. Based on the following data, test this claim. Data: 42 88 99 51 57 78 92 46 57

Ex #5 t test – placement scores H0: µ 75 fill in the correct hypothesis here H1: µ 75 α = 0.05 t = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #6- placement scores The head of the tutoring department claims that the average placement score is below 80. Based on the following data, test this claim. Data: 42 88 99 51 57 78 92 46 57

Ex #6– t example H0: µ 80 (fill in the correct hypotheses here) α = 0.05 t = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #7- salaries– t A national study shows that nurses earn $40,000. A career director claims that salaries in her town are higher than the national average. A sample provides the following data: 41,000 42,500 39,000 39,999 43,000 43,550 44,200

Ex #7- salaries H0: µ 40000 (fill in the correct hypotheses here) α = 0.05 t = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Traditional Critical Value Approach Redo Example #1 Recall that I claimed that my bag of numbers had a mean µ = 100 and a standard deviation =21.9. Test this hypothesis if your sample size n= 20 and your sample mean x-bar was 90.

Ex#1 redone with CV Z = = CV 4. Test conclusion α = 0.05 Z = = CV 4. Test conclusion If p-value ≤ α, then test value is in RR, and we reject H0 and say that the data are significant at level α If p-value > α, then test value is not in RR, and we do not reject H0 5. Interpretation of test results

Ex #3 redone with CV A group has a mean cholesterol of 220. The data is normally distributed with σ= 15 After a new drug is used, test the claim that it lowers cholesterol. Data: n=30, sample mean= 214.

Ex#3- 5 steps- done with CV H1: µ 220 (fill in) α = 0.05 Z = = CV 4. Test conclusion If p-value ≤ α, then test value is in RR, and we reject H0 and say that the data are significant at level α If p-value > α, then test value is not in RR, and we do not reject H0 5. Interpretation of test results

9.3 Testing Proportion p Recall confidence intervals for p: ± z

Hypothesis tests for proportions Null hypothesis H0, alternative hypothesis H1, and preset α 2. Test statistic and sampling distribution P-value and/or critical value z= = 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results 2-tailed ex H0: p= .5 H1: p ≠ .5 α = 0.05 Left tail ex H0: p = .7 H1: p < .7 Right tail ex H0: p = .2 H1: p > .2

Ex #8- proportion who like job The HR director at a large corporation estimates that 75% of employees enjoy their jobs. From a sample of 200 people, 142 answer that they do. Test the HR director’s claim.

Ex #8 H0: p=.75 (fill in hypothesis) H1: p α = Null hypothesis H0, alternative hypothesis H1, and preset α H0: p=.75 (fill in hypothesis) H1: p α = Test statistic and sampling distribution Z = = 3. P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #9 Previous studies show that 29% of eligible voters vote in the mid-terms. News pundits estimate that turnout will be lower than usual. A random sample of 800 adults reveals that 200 planned to vote in the mid-term elections. At the 1% level, test the news pundits’ predictions.

Ex #9 H0: p (fill in hypothesis) H1: p α = Null hypothesis H0, alternative hypothesis H1, and preset α H0: p (fill in hypothesis) H1: p α = Test statistic and sampling distribution Z = = 3. P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results