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11.1 – Significance Tests: The basics. Inference:. to assess the evidence provided by the sample to claim information about the population. Hypothesis:. A claim made about a population parameter, and sample data is gathered to determine whether the hypothesis is true. Null Hypothesis:.
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Inference: to assess the evidence provided by the sample to claim information about the population.
Hypothesis: A claim made about a population parameter, and sample data is gathered to determine whether the hypothesis is true.
Null Hypothesis: The statement being tested. We believe this to be true until we get evidence against it. NOTATION:
Alternate Hypothesis: Statement we hope or suspect is true instead of the null hypothesis NOTATION:
Test Statistic: A sample statistic that is computed from the data. It helps us to make a statistical decision. Do we have enough evidence to reject the null hypothesis or not? test statistic = Z =
p-value: • This value measures how much evidence you have against the null hypothesis. • Small p-values indicate the outcome measured from the sample data is unlikely given the null hypothesis is true. It provides strong evidence against your null hypothesis.
Statistically Significant: An event unlikely to occur by chance. If your p-value is small, then it is statistically significant. It is called alpha, .
Significance Level: The decisive p-value we fix in advance. This states when the null hypothesis should be rejected. This level is compared to the p-value. Common levels of rejection are =0.10, =0.05, and =0.01. p < , then reject the null p, then accept the null
Conditions: SRS Normality Independence
Example #1 – State the notation for the null and alternative hypothesis • Suppose we work in the quality control department of Ruffles Potato Chips. The quality control manager wants us to verify that the filling machine is calibrated properly. We wish to determine if the mean amount of chips in a bag is different from the advertised 12.5 ounces. The company is concerned if there are too many or too few chips in the bag.
Example #1 – State the notation for the null and alternative hypothesis b. According to the US Department of Agriculture, the mean farm rent in Indiana was $89 per acre in 1995. A researcher for the USDA claims that the mean rent has decreased since then. He randomly selected 50 farms from Indiana and determined the mean farm rent to be $67.
Example #1 – State the notation for the null and alternative hypothesis c. Researchers claim to have found a brain protein that blocks the craving for fatty food and therefore, increases the loss of body fat. To test this theory, 100 people are treated with protein and the reduction in body fat is measured.
Example #2 At the bakery where you work, loaves of bread are supposed to weigh 1 pound. From experience, the weights of loaves produced at the bakery follow a Normal distribution with standard deviation = 0.13 pounds. You believe that new personnel are producing loaves that are heavier than 1 pound. As supervisor of Quality Control, you want to test your claim at the 5% significance level. You weigh 20 loaves and obtain a mean weight of 1.05 pounds. a. Identify the parameter of interest. State your null and alternative hypotheses. = True mean weight of loaves of bread
Example #2 At the bakery where you work, loaves of bread are supposed to weigh 1 pound. From experience, the weights of loaves produced at the bakery follow a Normal distribution with standard deviation = 0.13 pounds. You believe that new personnel are producing loaves that are heavier than 1 pound. As supervisor of Quality Control, you want to test your claim at the 5% significance level. You weigh 20 loaves and obtain a mean weight of 1.05 pounds. b. Verify the conditions are met. SRS (must assume) (yes, pop. is approx normal, therefore, so is sample dist) Normality Independence (There are more than 200 loaves of bread)
Example #2 At the bakery where you work, loaves of bread are supposed to weigh 1 pound. From experience, the weights of loaves produced at the bakery follow a Normal distribution with standard deviation = 0.13 pounds. You believe that new personnel are producing loaves that are heavier than 1 pound. As supervisor of Quality Control, you want to test your claim at the 5% significance level. You weigh 20 loaves and obtain a mean weight of 1.05 pounds. c. Calculate the test statistic and the P-value. Illustrate using the graph provided.
P(Z > 1.72) = 1 – P(Z < 1.72) =
P(Z > 1.72) = 1 – P(Z < 1.72) = 1 – 0.9573 = 0.0427
Example #2 At the bakery where you work, loaves of bread are supposed to weigh 1 pound. From experience, the weights of loaves produced at the bakery follow a Normal distribution with standard deviation = 0.13 pounds. You believe that new personnel are producing loaves that are heavier than 1 pound. As supervisor of Quality Control, you want to test your claim at the 5% significance level. You weigh 20 loaves and obtain a mean weight of 1.05 pounds. d. State your conclusions clearly in complete sentences. I would reject the null hypothesis at the 0.05 level. I believe that the workers are making the loaves heavier.
Steps to Hypothesis Testing: PHANTOMS P: Parameter of interest H: Hypothesis A: Assumptions N: Name of Test T: Test Statistic O: Obtain P-Value M: Make a Statistical Decision S: Summary in context of problem.
One-Sample Z-Test: Testing the mean when is known.
Calculator Tip: Z-Test Stat – Tests - ZTest
Example #1 An energy official claims that the oil output per well in the US has declined from the 1998 level of 11.1 barrels per day. He randomly samples 50 wells throughout the US and determines that the mean output to be 10.7 barrels per day. Assume =1.3 barrels. Test the researchers claim at the =0.05 level. Mean oil output per well in the US P: H:
A: (says so) SRS Normality (n 30, so by the CLT, approx normal) Independence (Safe to assume more than 500 wells in the US) ZTest N:
O: P(Z < -2.176) =
O: P(Z < -2.176) = 0.0146
M: < 0.0146 0.05 Reject the Null
S: There is enough evidence to reject the claim that the average oil output per well in the US is 11.1 barrels per day.
Example #2 The average daily volume of Dell computer stock in 2000 was 31.8 million shares with a standard deviation of 14.8 million shares according to Yahoo! A stock analyst claims that the stock volume in 2001 is different from the 2000 level. Based on a random sample of 35 trading days in 2001, he finds the sample mean to be 37.2 million shares. Test the analyst’s claim at the =0.01 level. Mean volume of Dell computer stock P: H:
A: (says so) SRS Normality (n 30, so by the CLT, approx normal) Independence (Safe to assume more than 350 trading days) ZTest N:
O: 2[ P(Z < -2.16)] =
O: 2[ P(Z < -2.16)] = 2[ 0.0154] = 0.0308
M: > 0.0308 0.01 Accept the Null
S: There is not enough evidence to claim that the average daily volume of Dell stock is different from 31.8 million shares.
Duality of Confidence Intervals and Hypothesis Testing If the confidence interval does not contain μo, we have evidence that supports the alternative hypothesis, thus we reject the null hypothesis at the level. Note: The Confidence Interval matches the two-tailed test only!
Example #3 The average daily volume of Dell computer stock in 2000 was 31.8 million shares with a standard deviation of 14.8 million shares according to Yahoo! A stock analyst claims that the stock volume in 2001 is different from the 2000 level. Based on a random sample of 35 trading days in 2001, he finds the sample mean to be 37.2 million shares. Test the analyst’s claim at the =0.01 level. • What was your conclusion from this hypothesis test in Example #2? To not reject 31.8 million shares per day
b. Construct a 99% confidence interval for the true average daily volume of Dell Computer stock in 2001. Note: We already did P and A N: Z-Interval
I am 99% confident the true mean daily Dell volume stock is between 30.756 and 43.644 million shares. C: c. Does this interval reaffirm your statistical decision from the hypothesis test? Explain. Yes, 31.8 is in the interval, so can’t assume it is different
Example #4 Does marijuana use affect anger expression? Assume for all non-users, the mean score on an anger expression scale is 41.5 with a standard deviation of 6.05. For a random sample of 47 frequent marijuana users, the mean score was 44. • Test the claim that marijuana affects the expression of anger at the =0.05 level. True mean anger expression for marijuana users P: H:
A: (says so) SRS Normality (n 30, so by the CLT, approx normal) Independence (Safe to assume more than 470 marijuana users) ZTest N:
O: 2[ P(Z < -2.83)] =