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TOPIC 8 Inferences on a Population Mean

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Road Map Estimation Statistical Methods Descriptive Statistics Statistical Inference Decision Making

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Mean, , is unknown Population I am 95% confident that is between 40 & 60. Mean X = 50 Random Sample Sample Estimation Process

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Point Estimates Estimating Population Parameters Using Sample Statistics Mean ( µ ) Standard Deviation ( ) S Proportion ( p ) These are single value estimates (point estimates) They do not tell us how close our estimate is to the actual unknown parameters As the sample statistic varies from sample to sample, an interval based on the value of the sample statistics provides an estimate of the population parameter.

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Sample statistic (point estimate) Confidence interval Confidence limit (lower) Confidence limit (upper) A probability that the population parameter falls somewhere within the interval. Key Elements of Interval Estimation

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/2 x x = 1 - X _ _ Sampling Distribution of Sample Mean Large number of intervals (1 – α)% of intervals contain μ α% do not Intervals extend from X – Zσ X to X + Zσ X Intervals and Confidence Level

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Interval Estimates By taking all possible samples of n and compute their sample means, you’ll see that 95% of the intervals will include the population mean, and only 5% of them will not. In other words, you have 95% confidence that the population mean is somewhere in your interval. This tells us how close the estimates are to the true population parameter

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Confidence Level The common level of confidence used are 90%, 95% or 99% The level of confidence is symbolized by (1 – α) 100%, where α is the proportion of the tails, upper (α/2) and lower (α/2), outside the confidence interval. Confidence Level (1- α)100% αα/2 Critical Value Z α/2 90% 95% 99% 0.1 0.05 0.01 0.05 0.025 0.005 1.645 1.960 2.575

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1. Data dispersion Measured by Intervals extend from X – Z X to X + Z X 3.Level of confidence (1 – ) Affects Z 2.Sample size Factors Affecting Interval Width

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Unknown Known Interval Estimation Assumptions: Population follows a normal distribution or Sample size is large (n ≥ 30) The value of Z α/2 changes according to the confidence level (90%,95% or 99%) Confidence Interval for μ (σ Known)

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You’re a Q/C inspector for Gallo. The for 2-liter bottles is.05 liters. A random sample of 100 bottles showed x = 1.99 liters. What is the 90% confidence interval estimate of the true mean amount in 2-liter bottles? 2 liter ExampleExample

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Example Solution (Look up the table or find using the normal table)

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ExerciseExercise The quality control manager at a light bulb factory needs to estimate the mean life of a large shipment of light bulbs. The standard deviation is 100 hours. A random sample of 64 light bulbs indicated a sample mean life of 350 hours. Construct a 95% confidence interval estimate of the population mean life of light bulbs in this shipment. Do you think that the manufacturer has the right to state that the light bulbs last an average of 400 hours? Explain. Must you assume that the population of light bulb life is normally distributed? Explain.

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a) b) No. c) Not necessary Exercise Solution (Look up the table or find using the normal table)

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Interval Estimation Unknown Known The sample standard deviation S will be a good estimator. Then the confidence interval is approximately Confidence Interval for μ (σ Unknown) Assumptions: Sample size is large (n ≥ 30)

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ExerciseExercise A stationary store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 100 greeting cards indicates a mean value of $2.55 and a standard deviation of $0.44 Construct a 95% confidence interval estimate of the mean value of all greeting cards in the store’s inventory. Suppose there were 2,500 greeting cards in the store’s inventory. How are the results above useful in assisting the store owner to estimate the total value of her inventory?

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a) b) Exercise Solution (Look up the table or find using the normal table)

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Confidence Interval for μ (σ Unknown) What if the sample size is small (less than 30). Instead of using the standard normal statistic which requires knowledge of a good approximation of σ, we define and use a statistic, which is known as Student’s t distribution. If the random variable X is normally distributed, then the following test statistic has a t distribution with n – 1 degrees of freedom ( df ) Please notice that some books, including our textbook, use student’s t statistic even for large sample size ( n ≥ 30). This applies also for next topic, hypothesis testing!

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Z t 0 t (df = 5) Standard Normal t (df = 13) Bell-Shaped Symmetric ‘Fatter’ Tails Student’s t Distribution

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1. Number of observations that are free to vary after sample statistic has been calculated 2. Example Sum of 3 numbers is 6 X 1 = 1 (or any number) X 2 = 2 (or any number) X 3 = 3 (cannot vary) Sum = 6 degrees of freedom = n - 1 = 3 - 1 = 2 Degrees of Freedom (df)

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d.f 0.250.10.050.025 1.00003.07776.313712.7062 0.81651.88562.92004.3027 0.76491.63772.35343.1824 0.74071.53322.13182.7765 0.72671.47592.01502.5706 1 2 3 4 5 6 0.71761.43981.94322.4469 t - Table Degrees of freedom, df = n - 1 Match ‘df’ and upper tail area, α/2 For example, ‘t’ value with 90% confidence level and sample size, n = 6 α = 10%, α/2 = 5% = 0.05 df = n - 1 = 6 - 1 = 5 Match (5, 0.05) in the table ‘t α/2 ’ value = 2.0150

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Confidence Interval for μ (σ Unknown) Assumptions: σ unknown Population follows a normal distribution Sample size is small (n < 30)

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You’re a time study analyst in manufacturing. You’ve recorded the following task times (min.): 3.6, 4.2, 4.0, 3.5, 3.8, 3.1. What is the 90% confidence interval estimate of the population mean task time? ExampleExample

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n = 6, df = n - 1 = 6 - 1 = 5, t.05 = ±2.015 Example Solution (Look up the table or find using the t table)

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ExerciseExercise The following data represent the bounced check fee, in dollars, charged by a sample of 23 banks for direct- deposit customers who maintain a $100 balance: 26 28 20 20 21 22 25 25 18 25 15 20 18 20 25 25 22 30 30 30 15 20 29 Construct a 90% confidence interval for the population mean bounced check fee. What assumption is required for the interval to be valid? Is it reasonably satisfied?

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n = 23, df = n - 1 = 23 - 1 = 22 t.05 = ±1.717 Exercise Solution (Look up the table or find using the t table) b) It is assumed that the population distribution follows a normal distribution a)

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SE = Sampling Error I don’t want to sample too much or too little! Sample Size for Estimating µ Sample standard deviation S from prior sampling would be a good estimator. Please notice that the textbook that we use has different formula to determine the sample size. L is interval length

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What sample size is needed to be 90% confident the mean is within 5? A pilot study suggested that the standard deviation is 45. ExampleExample

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Road Map Estimation Statistical Methods Descriptive Statistics Statistical Inference Decision Making

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Hypothesis Testing Process Population I believe the population mean age is 50 (hypothesis). Reject hypothesis! Not close. Mean X = 20 Random sample

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What is a Hypothesis? A Statement/s or claim about a population parameter ( μ, p, σ), developed for the purpose of testing. Hypothesis statement/s are made before analysis. We claim that the average mileage is 10 km/l There are two elements of hypothesis: Null hypothesis Alternative hypothesis

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Null and Alternative Hypothesis A Null Hypothesis is a statement that nothing unusual occurs or will occur. We begin with the assumption it is true Designated H 0 Always has equality sign: = Alternative Hypothesis is the opposite of null hypothesis. Alternative hypothesis is a statement that something unusual occurs or will occur: Designated H 1 or H a Always has inequality sign: ≠,

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Example: Test 1 “Test that the population mean is not 5” 1)State the question statistically: μ ≠ 5 2)State the opposite statistically: μ = 5 Must be mutually exclusive & exhaustive 3)Select the null hypothesis: H 0 : μ = 5 4)Hence the alternate hypothesis is: H a : μ ≠ 5 (two tailed test)

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Example: Test 2 “Test that the population mean is less than 5” 1)State the question statistically: μ < 5 2)State the opposite statistically: μ ≥ 5 Must be mutually exclusive & exhaustive 3)Select the null hypothesis: H 0 : μ ≥ 5 4)Hence the alternate hypothesis is: H a : μ < 5 (one tailed test / lower tailed test) “Test that the population mean is greater than 5” Null hypothesis: H 0 : μ ≤ 5 Alternate hypothesis: H a : μ > 5 (one tailed test / upper tailed test)

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Critical Value and Rejection Region The dividing point (critical value) between the region where the null hypothesis is rejected and the region where it is not rejected 1)based on selected α value (level of significance) 2)and the type of test statistic Z or t 3)and whether a one or two tail test is used. 1/2 Z or t Rejection Region Rejection Region Nonrejection Region 1 – Critical value (Z α/2 or t α/2 ) Critical value (-Z α/2 or –t α/2 )

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1) Level of Significance ( α ) Probability of rejecting a true null hypothesis Represented by ‘α’ (alpha) Selected by researcher at start Typical values are 0.01, 0.05, 0.10 or (0.01 – 0.10) Same as confidence level of (1 - α) 100% For example, α = 0.01 (1%) means 99% confidence level

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2) Type of Test Statistic: Z or t ? Estimation Z – Test 1 or 2 tailed test One Sample of Population σ Known σ Unknown t – Test 1 or 2 tailed test

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3) Whether One or Two Tailed Test ? H 0 : 5 H a : 5 For a two-tailed test (H a : 5) both the upper and lower tails marked by the critical values are the rejection region for H 0 If the calculated test statistic (Z or t) lies in the rejection region, the null hypothesis (H 0 ) is rejected. Otherwise, accept H 0 Example 1: HoHo Value 1/2 Sample Statistic Rejection Region Rejection Region Nonrejection Region 1 – Critical value (Z α/2 or t α/2 ) Critical value (-Z α/2 or –t α/2 ) Two Tailed Test

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3) Whether One or Two Tailed Test ? 0 Reject H 0 Critical value (Z α or t α ) H 0 : 5 H a : 5 For a one/upper-tailed test (H a : > 5) the rejection region is the right-hand tail marked by a positive critical value. If the calculated test statistic (Z or t) lies in the rejection region, the null hypothesis (H 0 ) is rejected. Otherwise, accept H 0 Example 2: Accept H 0 HoHo Value Sample Statistic Rejection Region Nonrejection Region 1 – Critical value (Z α or t α ) One/Upper Tailed Test

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3) Whether One or Two Tailed Test ? 0 Critical value (-Z α or –t α ) Reject H 0 H 0 : 5 H a : 5 For a one/lower-tailed test (H a : < 5) the rejection region is the left-hand tail marked by a negative critical value. If the calculated test statistic (Z or t) lies in the rejection region, the null hypothesis (H 0 ) is rejected. Otherwise, accept H 0 Example 3: Accept H 0 HoHo Value Sample Statistic Rejection Region Nonrejection Region 1 – Critical value (-Z α or –t α ) One/Lower Tailed Test

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Steps in Hypothesis Testing 1)Set up the null and alternative hypothesis H 0, H a based on the research question 2)Decide the test to perform (Z or t) depending on whether is know or unknown 3)Determine the critical value for a given α 4)Compute the test statistic (Z or t) depending on the type of test used 5)Make decision using the rule (reject / accept H 0 )

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Z - Test of Hypothesis for μ Unknown Known Hypothesis Testing Z Test Assumptions: The population follows a normal distribution or The sample size is sufficiently large (n ≥ 30).

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Test Statistic: Z Test Alternative Hypothesis Lower-Tailed Z < -Z α Upper-Tailed Z > Z α Two-Tailed Z Z α/2 α = 0.10 α = 0.05 α = 0.01 Z < -1.280 Z < -1.645 Z < -2.330 Z > 1.280 Z > 1.645 Z > 2.330 Z 1.645 Z 1.960 Z 2.575 Rejection Regions for Common Values of α

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How to Find Critical Z Two Tailed Test.500 -.025.475 Z 0 = 1 What is Z given =.05? =.025 Z.05.07 1.64505.4515.4525 1.7.4599.4608.4616 1.8.4678.4686.4693.4744.4756.06 1.9.4750 Standardized Normal Probability Table (Portion) 1.96 -1.96 Z = 1.96

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One Tailed Test.500 -.025.475 Z 0 = 1 What is Z given =.025? =.025 Z.05.07 1.64505.4515.4525 1.7.4599.4608.4616 1.8.4678.4686.4693.4744.4756.06 1.9.4750 Standardized Normal Probability Table (Portion) 1.96 -1.96 Z = 1.96 How to Find Critical Z

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Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 15 grams. Test at the.05 level of significance. 368 gm.368 gm. Example: Two Tailed Test

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H 0 : H a : , n Critical Value(s): Test Statistic: Z = 1.5 is in the non rejection region Decision: Conclusion: = 368 368.05 25 Z 01.96-1.96.025 Reject H 0 0.025 Do not reject at =.05 No evidence average is not 368 Example Solution

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You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. At the.01 level of significance, is there evidence that the miles per gallon is at least 32? Example: One Tailed Test

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H 0 : H a : = n = Critical Value(s): Test Statistic: Z = -2.65 is in the rejection region Decision: Conclusion: = 32 < 32.01 60 Z 0-2.33.01 Reject Reject at =.01 There is evidence average is less than 32 Example Solution

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ExercisesExercises 1)The quality control manager believes that the lifetime of bulbs follows a normal distribution with a mean of 200 hours and a standard deviation of 20 hours. A sample of 100 bulbs showed a sample mean of 195 hours. The manager wants to test whether the mean lifetime of all bulbs is 200 at a 1% level of significance. 2)A company that manufactures chocolate bars is particularly concerned that the mean weight of a chocolate bar not be greater than 6.03 ounces. Past experience allows you to assume that the standard deviation is 0.02 ounces. A sample of 50 chocolate bars is selected, and the sample mean is 6.034 ounces. Using the α = 0.01 level of significance, is there evidence that the population mean weight of chocolate bars is greater than 6.03 ounces?

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H 0 : H a : = n = Critical Value(s): Test Statistic: Z = -2.5 is in the rejection region Decision: Conclusion: = 200 < 200.01 100 Z 0-2.33.01 Rejection Region Reject H 0 at =.01 There is evidence average is less than 200 hours Exercise 1 Solution

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H 0 : H a : = n = Critical Value(s): Test Statistic: Z = 1.41 is in the non rejection region Decision: Conclusion: ≤ 6.03 > 6.03.01 50 Accept H 0 at =.01 There is no evidence average is greater than 6.03 ounces Exercise 2 Solution Rejection Region.01 2.33

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Z - Test of Hypothesis for μ Interval Estimation Unknown Known Small Sample t Test Large Sample Z Test Test Statistic:

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ExampleExample A random sample of 64 observations produced the following summary statistics: and Test the null hypothesis that the mean of the population is 0.36 against the alternative hypothesis, μ < 0.36. Use α = 0.1. Test the null hypothesis that the mean of the population is 0.36 against the alternative hypothesis, μ ≠ 0.36. Use α = 0.1. Inteprete the result.

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H 0 : H a : n Critical Value(s): Test Statistic: Z = -1.6 is in the rejection region Decision: Conclusion: = 0.36 0.36 0.1 64 Z 0-1.28 0.1 Reject H 0 at = 0.1 There is evidence average is less than 0.36 Example Solution (1) Rejection Region

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H 0 : H a : , n Critical Value(s): Test Statistic: Z = -1.6 is in the non rejection region Decision: Conclusion: = 0.36 0.36 0.1 64 Do not reject H 0 at = 0.1 No evidence average is not 0.36 Example Solution (2) Z 01.645-1.645.05 Rejection Region Rejection Region

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Interval Estimation Unknown Known Small Sample t Test Large Sample Z Test t - Test for Mean (μ), σ Unknown

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1. Assumptions Population is normally distributed If not normal, only slightly skewed & small sample (n 30) taken 2. Parametric test procedure 3. t test statistic t - Test for Mean (μ), σ Unknown

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t 0 Given: n = 3; =.10 /2 =.05 df = n - 1 = 2 v t.10 t.05 t.025 13.0786.314 12.706 21.8862.9204.303 31.6382.3533.182 Critical Values of t Table (Portion) 2.920-2.920 How to Find Critical t Two Tailed Test

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You’re a marketing analyst for Wal- Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was: 8 11 0 4 7 8 10 5 8 3 At the.05 level of significance, is there evidence that the average bear sales per store is different from 5 ($ 00)? the average bear sales per store is more than 5 ($ 00)? ExampleExample

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H0: Ha: = 0.025 df = Critical Value(s): Test Statistic: t = 1.31 is in the non rejection region Decision: Conclusion: = 5 5.05 10 - 1 = 9 Do not reject H 0 at =.05 There is no evidence average is different from 5 Example Solution (1) t 02.262-2.262.025 Rejection Region Rejection Region

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H0: Ha: = df = Critical Value(s): = 5 > 5.05 10 - 1 = 9 t 01.833.05 Example Solution (2) Test Statistic: t = 1.31 is in the non rejection region Decision: Conclusion: Do not reject H 0 at =.05 There is no evidence average is more than 5 Rejection Region

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Exercise: A particular branch of a well-known bank stores enough money during the weekends to satisfy its customer’s needs. The expected average withdrawal during the weekend is $500. When they looked at a sample of the last 16 weekend transactions, they found the average withdrawal to be $540 with a standard deviation of $70. At =.05, is there evidence that the mean withdrawal has increased during weekends? It is assumed that the population follows a normal distribution. ExerciseExercise

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H0: Ha: = df = Critical Value(s): = 500 > 500.05 n – 1 = 16 - 1 = 15 t 01.753.05 Exercise Solution Test Statistic: t = 2.28 is in the non rejection region Decision: Conclusion: Reject H 0 at =.05 There is evidence average is more than 500 Rejection Region

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Observed Significance Level (p -Values) The p-value is the probability of getting a test statistic equal to or more extreme than the sample result, give that the null hypothesis, H 0, is true. The p-value, often referred to as the observed level of significance, is the smallest value of the significance level at which H 0 can be rejected. The decision rules for rejecting H 0 in the p-value approach are If the p-value is less than , the null hypothesis is rejected If the p-value is greater than or equal to , the null hypothesis is not rejected

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Finding the p - Values Z p-value = Area p-value = 2 × Probability ( Z ≥ ‘absolute value’ of the computed test statistic value) p-value = Probability ( Z ≥ the computed test statistic value) p-value = Probability ( Z ≤ the computed test statistic value) Z p-value = Area Z Area p-value = 2 × Area

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You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. What is the value of the observed level of significance (p-Value)? ExampleExample

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H 0 : H a : σ n Critical Value(s): Test Statistic: ≥ 32 32 3.8 60 Z 0 -2.65 p-value Example Solution Rejection Region p-Value is P(Z -2.65) =.004 Since p-Value < ( =.01) then Reject H 0.

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ExerciseExercise The quality control manager believes that the lifetime of bulbs follows a normal distribution with a mean of 200 hours and a standard deviation of 20 hours. A sample of 100 bulbs showed a sample mean of 195 hours. The manager wants to test whether the mean lifetime of all bulbs is 200 at a 1% level of significance. Use the p-value approach.

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H 0 : H a : σ n Critical Value(s): Test Statistic: ≥ 200 200 20 100 Z 0 -2.50 p-value Exercise Solution Rejection Region p-Value is P(Z -2.50) =.0062 Since p-Value < ( =.01) then Reject H 0.

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Risks in Using Hypothesis Testing Statistical Decision Actual Situation H 0 TrueH 0 False Do not reject H 0 Correct Decision Confidence = (1-α) Type II error P (Type II error) = β Reject H 0 Type I error P (Type I error) = α Correct Decision Power = (1 – β) Type I error has serious consequences

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You can’t reduce both errors simultaneously! and Have an Inverse Relationship and Have an Inverse Relationship

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