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Lesson 11 - 2 Carrying Out Significance Tests
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Lesson 11 - 2. Carrying Out Significance Tests. Knowledge Objectives. Identify and explain the four steps involved in formal hypothesis testing. Construction Objectives. Using the Inference Toolbox, conduct a z test for a population mean . - PowerPoint PPT Presentation

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• Lesson 11 - 2Carrying Out Significance Tests

• Knowledge ObjectivesIdentify and explain the four steps involved in formal hypothesis testing.

• Construction ObjectivesUsing the Inference Toolbox, conduct a z test for a population mean.

Explain the relationship between a level two-sided significance test for and a level 1 confidence interval for .

Conduct a two-sided significance test for using a confidence interval.

• VocabularyHypothesis a statement or claim regarding a characteristic of one or more populations

Hypothesis Testing procedure, base on sample evidence and probability, used to test hypotheses

Null Hypothesis H0, is a statement to be tested; assumed to be true until evidence indicates otherwise

Alternative Hypothesis H1, is a claim to be tested.(what we will test to see if evidence supports the possibility)

Level of Significance probability of making a Type I error,

• Inference ToolboxTo test a claim about an unknown population parameterStep 1: HypothesesIdentify population of interest and parameterState null and alternative hypotheses Step 2: ConditionsCheck 3 conditions (SRS, Normality, Independence); if met, then continue (or proceed under caution if not)Step 3: CalculationsState test or test statisticUse calculator to calculate test statistic and p-valueStep 4: InterpretationConclusion, connection and context about P-value and the hypotheses

• Z Test for a Population MeanNote: not usually seen in AP world (t-test more common)

• Hypothesis Testing ApproachesClassicalLogic: If the sample mean is too many standard deviations from the mean stated in the null hypothesis, then we reject the null hypothesis (accept the alternative)

P-ValueLogic: Assuming H0 is true, if the probability of getting a sample mean as extreme or more extreme than the one obtained is small, then we reject the null hypothesis (accept the alternative).

Confidence IntervalsLogic: If the sample mean lies in the confidence interval about the status quo, then we fail to reject the null hypothesis

• Critical RegionsClassical Approach

Reject null hypothesis, ifLeft-TailedTwo-TailedRight-Tailedz0 < - zz0 < - z/2orz0 > z /2z0 > z

• P-Value is the area highlightedP-Value Approach

Reject null hypothesis, ifP-Value <

• Confidence Interval ApproachLower BoundUpper Bound0

Reject null hypothesis, if0 is not in the confidence interval

• Example 1Assume that cell phone bills are normally distributed. A simple random sample of 12 cell phone bills finds x-bar = \$65.014. The mean in 2004 was \$50.64. Assume = \$18.49. Test if the average bill is different today at the = 0.05 level. Use each approach.Step 1: Hypothesis H0: = \$50.64 (mean cell phone bill is unchanged) Ha: \$50.64 (mean cell phone bill has changed)

Step 2: Conditions SRS: stated in problem Normality: population is normally distributed Independence: far more than 120 cell phones

• Example 1: Classical ApproachA simple random sample of 12 cell phone bills finds x-bar = \$65.014. The mean in 2004 was \$50.64. Assume = \$18.49. Test if the average bill is different today at the = 0.05 level. Use the classical approach.Using alpha, = 0.05 the shaded region are the rejection regions. The sample mean would be too many standard deviations away from the population mean. Since z0 lies in the rejection region, we would reject H0.

• Example 1: P-Value A simple random sample of 12 cell phone bills finds x-bar = \$65.014. The mean in 2004 was \$50.64. Assume = \$18.49. Test if the average bill is different today at the = 0.05 level. Use the P-value approach. X-bar 65.014 50.64 14.374Z0 = --------------- = ---------------------- = ------------- = 2.69 / n 18.49/12 5.3376not equal two-tailed-Z0 = -2.69 The shaded region is the probability of obtaining a sample mean that is greater than \$65.014; which is equal to 2(0.0036) = 0.0072. Using alpha, = 0.05, we would reject H0 because the p-value is less than .P( z < Z0 = -2.69) = 0.0036 (double this to get p-value because its two-sided!)

• Using Your Calculator: Z-TestFor classical or p-value approachesPress STATTab over to TESTSSelect Z-Test and ENTERHighlight StatsEntry 0, , x-bar, and n from summary statsHighlight test type (two-sided, left, or right)Highlight Calculate and ENTERRead z-critical and/or p-value off screenFrom previous problem: z0 = 2.693 and p-value = 0.0071

• Example 1: Confidence IntervalA simple random sample of 12 cell phone bills finds x-bar = \$65.014. The mean in 2004 was \$50.64. Assume = \$18.49. Test if the average bill is different today at the = 0.05 level. Use confidence intervals.The shaded region is the region outside the 1- , or 95% confidence interval. Since the old population mean lies outside the confidence interval, then we would reject H0.Confidence Interval = Point Estimate Margin of Error = x-bar Z/2 / n = 65.014 1.96 (18.49) / 12 = 65.014 10.4617Zc (/2) = 1.96

• Using Your Calculator: Z-IntervalPress STATTab over to TESTSSelect Z-Interval and ENTERHighlight StatsEntry , x-bar, and n from summary statsEntry your confidence level (1- )Highlight Calculate and ENTERRead confidence interval off of screenIf 0 is in the interval, then FTRIf 0 is outside the interval, then REJFrom previous problem: u0 = 50.64 and interval (54.552, 75.476)Therefore Reject

• Example 2National Center for Health has that the mean systolic blood pressure for males 35 to 44 years of age is 128. The medical director for a company examines the medical records of 72 male executives in the age group and finds that their mean blood pressure is 129.93. Is there evidence to support that their blood pressure is different?Step 1: Hypothesis H0: = 128 (younger male executives mean blood pressure is 128) Ha: 128 (their blood pressure is different than 128)

Step 2: Conditions SRS: possible issue, but selected from free annual exams Normality: sample size large enough for CLT to apply Independence: have to assume more than 720 young male executives in the company (large company!!)

• Example 2: P-Value Step 3: Calculations:

Step 4: Interpretation: X-bar 129.93 128 1.93Z0 = --------------- = ---------------------- = ------------- = 1.092 / n 15/72 1.7678From calculator: z = 1.0918 p-value = 0.2749More than 27% of the time with a sample size of 72 from the general population of males in the 35-44 age group, we would get blood pressure values this extreme or more Fail to reject H0; not enough evidence to say that this companies executives differ from the general population.

• Example 3Medical director for a large company 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 -6. We take the population standard deviation to be =20. The director decides to use an =0.05 significance level.

• Example 3 contHypothesis: H0: Ha:

Conditions: 1:

2:

3: = 0 blood pressure is same < 0 Regime lowers blood pressure SRS -- stated in the problem statementNormality -- unknown underlying distribution, but large sample size of 50 says x-bar will be Normally distributed (CLT)Independence -- since sampling is w/o replacement; assume company has over 500 employees

• Example 3 contCalculations:

Interpretation:

x 0 -6 0Z0 = ----------- = ------------------ /n 20/50

= -2.12P-value = P(z < Z0) = P(z < -2.12) = 0.0170 (unusual !) P-value = 0.0170 so only 1.7% of the time could we get a more extreme value. Since this is less than = 0.05, we reject H0 and conclude that the mean difference in blood pressure is negative (so the regime may have worked!)

• Example 4The Deely lab analyzes specimens of a drug to determine the concentration of the active ingredient. The results are not precise and repeated measurements follow a Normal distribution quite closely. The analysis procedure has no bias, so the mean of the population of all measurements is the true concentration of the specimen. The standard deviation of this distribution was found to be =0.0068 grams per liter.

A client sends a specimen for which the concentration of active ingredients is supposed to be 0.86%. Deelys three analyses give concentrations of 0.8403, 0.8363, and 0.8447. Is their significant evidence at the 1% level that the concentration is not 0.86%? Use a confidence interval approach as well as z-test.

• Example 4 contHypothesis: H0: Ha:

Conditions: 1:

2:

3: = 0.86 grams per liter 0.86 grams per literSRS -- assume that each analyses represents an observation in a simple random sampleNormality -- stated in the problem that distribution is NormalIndependence -- assume each test is independent for the others

• Example 4 contCalculations:

Interpretation:

x 0 0.8404 0.86Z0 = ----------- = ------------------ /n 0.0068/3

= -4.99P-value = 2P(z < Z0) = 2P(z < -4.99) = 0.0004 (unusual !) P-value = 0.0004 so only .04% of the time could we get a more extreme value. Since this is less than = 0.01, we reject H0 and conclude that the mean concentration of active ingredients is not 0.86x-bar z* / n0.8404 2.576 (0.0068) / 30.8404 0.0101

(0.8303, 0.8505)

• Summary and HomeworkSummaryA hypothesis test of means compares whether the true mean is eitherEqual to, or not equal to, 0Equal to, or less than, 0Equal to, or more than, 0There are three equivalent methods of performing the hypothesis testThe classical approachThe P-value approachThe confidence interval approach

Homeworkpg 713 ; 11.35 - 40 X-bar Z0 = --------------- / n

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