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Hypothesis Testing: One Sample Mean or Proportion

Hypothesis Testing: One Sample Mean or Proportion. P - Values, Sections Section 10.3 Testing a Proportion, Section 10.6 Power of a Test, Section 10.7. P-value Approach. State null and alternative hypotheses H 0 :  = 8000 H 1 :   8000 (two sided test)

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Hypothesis Testing: One Sample Mean or Proportion

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  1. Hypothesis Testing: One Sample Mean or Proportion P - Values, Sections Section 10.3 Testing a Proportion, Section 10.6 Power of a Test, Section 10.7

  2. P-value Approach • State null and alternative hypotheses • H0:  = 8000 • H1:  8000 (two sided test) • Set level of significance, ⍺ = 0.05 • Do not find the critical values • Calculate appropriate test statistic • Z = (7814.1-8000)/94.28 = -1.97 PP 3

  3. P-value • The p-value tells us the probability of observing a test statistic as extreme or more extreme than the one we observed if the null hypothesis is true • In this example, the p-value tells us the probability of observing the test statistic of -1.97 or one more extreme if  = 8000 • Think more extreme as |1.97| PP 3

  4. Sampling Distribution of Sample Means, Normally Distributed .4756 8185.5 7814.1 -1.97 0 1.97 What is the p-value of |1.97|? • Need P(0 < z < 1.97) = 0.4756 • Standard normal table • In one tail = .5000 - .4756 = .0244 • In two tails = .0244*2 = .0488 • p-value = .0488 • This tells us that if the null hypothesis is true we would observe a test statistic as extreme as 1.97 in 4.88 samples out of 100 samples (4.88% of the time) .0244 .0244 Z PP 3

  5. Making the Statistical Decision • Decide whether the p-value is small and implies the sample mean is a rare event if the null hypothesis is true or whether the p-value is large and implies the sample mean could be expected to occur if the null hypothesis is true • Use our conventional levels of alpha (levels of significance) as benchmarks • Compare the p-value we calculate from the data with the level of significance PP 3

  6. Compare the p-value with ⍺ • If  = .05, then • If p-value < .05, reject the null hypothesis • If p-value ≥ .05, do not reject the null hypothesis • If  = .01, then • If p-value < .01, reject the null hypothesis • If p-value ≥ .01, do not reject the null hypothesis PP 3

  7. P-value and Critical Values - Consistent • If the test statistic (|1.97|) is greater than the critical value (|1.96|), the p-value (.0488) must be smaller than .05 • We will reject the null hypothesis whether we use a critical value or a p-value approach PP 3

  8. Observed Significance Level • A p-value is: • the exact level of significance of the test statistic • the smallest value a can be and still allow us to reject the null hypothesis • the amount of area left in the tail beyond the test statistic for a one-tailed hypothesis test or • twice the amount of area left in the tail beyond the test statistic for a two-tailed test. • the probability of getting a test statistic from another sample that is at least as far from the hypothesized mean as this sample statistic is PP 3

  9. One Tail Tests • The researcher is interested in whether the population parameter is lower (or higher) than the stated value in the null • Candy manufacturer claims there are 100 candies on average in each $1.00 bag • After sampling a number of bags, you are concerned that this claim may be an overstatement • Do we care if there are more than 100 candies in a bag on average? PP 3

  10. Example of Lower Tail Test • Form the alternative hypothesis first, since it embodies the challenge • H0: μ = 100 or H0: μ ≥ 100 (No evidence to reject the manufacturer’s claim) • The equality is always in the null • H1: μ < 100 (The manufacturer’s claim is rejected; Mean number of candies is less than 100) PP 3

  11. Lower Tail Test • The rejection region is in the lower tail of the distribution • All of ⍺ is placed in the one tail • Reject the null • If we observe a small sample mean that is unlikely to occur if μ = 100 Sampling Distribution under the null hypothesis reject Do not reject PP 3

  12. Example of Upper Tail Test • A company is thinking about setting up an on-site day-care program for its employees. The CEO has stated that she will do so only if more than 80% of the employees favor such a decision. • When will an action take place? • H0: π ≤ 0.80 • H1: π > 0.80 PP 3

  13. Two Sided Tests • Always use = and ≠ in the statistical hypotheses • Directionless in that the alternative hypothesis allows for either the greater than (>) or less than (<) possibility PP 3

  14. One tailed tests • Are always directional • The alternative hypothesis uses either the greater than (>) or less than (<) sign • Only used when the researcher • Knows for certain that the outcome of an experiment is going to occur only in one direction • Is only interested in one direction of the experiment • In one-tailed problems, the researcher is trying to demonstrate that something is • Older, younger, higher, lower, more, less, greater, less than, has increased, has decreased and so on • The directionality to be demonstrated is placed in the alternative hypothesis PP 3

  15. Problem – Lifetime of Light Bulbs • The advertising campaign of a light bulb manufacturer claims that the mean lifetime of their bulbs is 1600 hours • The known population standard deviation is  = 120 hours • In your job for Consumer Report you have been asked to check whether the company is making false claims • You obtain a sample of 100 bulbs • The mean lifetime of the sample is 1,570 hours • Let  = .05 • Sample means are normally distributed via the CLT PP 3

  16. Problem – Lifetime of Light Bulbs • H0: ≥ 1600 • The company’s claim appears valid • H1:  < 1600 • The company’s claim appears to be an overstatement • Find critical value Z.05 = -1.64 • Look for probability of .4500 in standard normal table • Look for probability of .05 in t table for infinite df • DR: if (Z test statistic ≥ -1.64) Do not reject H0 • Z = (1570 - 1600)/(120/100) = -2.50 PP 3

  17. Sampling Distributionof the sample mean under the null hypothesis ⍺ = 0.05 Do not reject 1570 1600 -2.50 -1.64 Z Lifetime of Light Bulbs: Critical Value Approach • DR: if (Z test statistic ≥ -1.64) do reject H0 • Calculate test statistic • Z = (1570 - 1600)/(120/100) = -2.50 • Reject the H0 • The company appears to be overstating the lifetime of its bulbs -1.64 PP 3

  18. Lifetime of Light Bulbs: p-value Approach • H0: ≥ 1600 (The company’s claim appears valid) • H1:  < 1600 (The company’s claim appears to be an overstatement) • Sample means are normally distributed via the CLT • Let ⍺ = 0.05 • Find test statistic: • Z = (1570 - 1600)/(120/100) = -2.50 PP 3

  19. -2.50 0 Lifetime of Light Bulbs: p-Value Approach • How likely are we to observe a test statistic of -2.50 or one more extreme, if the null hypothesis is true? • Standard Normal Table • P(0 < Z < 2.50) =.4938 • Calculate area in tail • .5000 - .4938 = .0062 • Compare with level of significance • .0062 < .05, reject null Sampling Distributionof the sample mean under the null hypothesis p-value = .0062 .4938 1570 1600 Z PP 3

  20. The t-Test Statistic • Consider the population standard deviation, , is unknown • Sampling distribution of Sample Means is normal because • Population follows the normal distribution • Or the sample size is greater than 30 (CLT) • How do we estimate the standard error of the sample means? • Substitute S, the sample standard deviation, for  PP 3

  21. T-test Problem: Soft Drink • A soft drink manufacturer wants to test whether the mean rating for a new flavor that it has just developed equals 60. From many previous studies it is known that its old flavor has a mean rating of 60. • If the mean rating for the new flavor is higher than 60, the manufacturer will substitute the new flavor for the old and stop producing the latter. • The manufacturer samples 16 people and obtains ratings on a scale of 1 to 90. The sample mean is 62.38 and the sample standard deviation is 7.6. • Should the manufacturer continue to produce the old flavor or should it shift production to the new flavor? • The ratings are normally distributed in the population. Let  = .05 PP 3

  22. T-test Problem • Where is the action? • If the mean rating for the new flavor is higher than 60, the manufacturer will substitute the new flavor for the old and stop producing the latter • H0: μ ≤ 60 (The new is the same or worse than the old. Do not switch) • H1: μ > 60 (The new is better than the old. Switch) • One sided, upper tail test PP 3

  23. Sampling Distributionof the sample mean under the null hypothesis 62.38 1.25 1.753 T-test Problem • Estimate the standard error • Calculate the test statistict15 = (62.38 - 60)/1.9 = 1.25 • Find the CV – t table • t15,.05 = 1.753 • Do not reject • Do not switch to new reject ⍺ = .05 Do not reject 60 Z PP 3

  24. Testing Hypotheses about the Population Proportion • Proportions are used for qualitative data • Observations fall into one category or another • Will you vote for the candidate: yes or no • Do you like the flavor: yes or no • We are interested in some characteristic of the observation • Population proportion is of interest • π = number of observations with characteristic/ population size = x/N PP 3

  25. Testing Hypotheses about the Population Proportion • Have the sample information • p = observations with characteristic/ sample size = x/n • If we had drawn a different sample, what might p in that sample be? • Need to understand the sampling distribution of the sample proportion • Can calculate exactly what the sampling distribution looks like PP 3

  26. The Sampling Distribution of the Sample Proportion • Suppose I have a large population • Fifty percent vote yes on an issue and fifty percent vote no • Interested in the proportion voting yes (a success) • Let π = 0.50 • Draw a sample of n = 5 • What would the distribution of sample proportions look like? • Answer lies with the binomial distribution • “What is the probability of observing exactly 0 (1, 2, 3, 4, 5) “successes” in 5 trials if the probability of success on a single trial is 0.5? PP 3

  27. The Binomial Distribution Use the Binomial Distribution tables in the appendix of Weiers to find the probability of observing exactly 0 successes in 5 trials, or 1 success in 5 trials and so forth. x = number of successes Proportion of successes Probability of x successes Using the binomial distribution, we can calculate the probability of observing certain sample proportions, given the population proportion is some specific value π = 0.50 and given the sample size, n = 5. PP 3

  28. Sampling Distribution of the sample proportion, p, when n = 5 and π = 0.50 P(x/n=p) = p The Sampling Distribution of the Sample Proportion • Is a binomial distribution • Not a continuous distribution • E(x/n) = E(p) = π • VAR(p) = π(1 - π)/n PP 3

  29. E(p) = π p Normal Distribution as an Approximation to the Binomial Sampling Distribution of p when n is “large” • Sampling distribution of the proportion will be approximately normal with mean π and standard deviation p • Conditions to be met • nπ > 5 when π 1/2 and n(1 - π) > 1/2 when π > 1/2 • Or sample size greater than 100 PP 3

  30. Hypotheses about the Population Proportion • Formulate the hypotheses • H0: π = π0 • H1: π ≠ π0 • or H1: π > π0 or H1: π < π0 • Set ⍺ and look-up Z critical value(s) in table • Form the rejection region assuming the null hypothesis is correct • Or set ⍺ and use the p-value approach PP 3

  31. Reject Reject E(p) = π0 p -Z⍺/2 Z⍺/2 Z Standard Error = Hypothesis about the Population Proportion Assuming the null hypothesis is correct, two sided test • Calculate the test statistic • Use π0in the standard error • If the test statistic is extreme, reject the H0 • DR: if (-Z⍺/2 ≤ Z test statistic ≥ Z⍺/2 ) do not reject H0 PP 3

  32. Proportion Problem • The manufacturer of an over-the-counter medicine claims that it is 90% effective in relieving an allergy for a period of 8 hours • In a sample of 200 persons who have the allergy, the medicine provided relief for 160 people • Determine whether the manufacturer’s claim is legitimate (overstating value?) • Let  = .01 PP 3

  33. Proportion Problem - OTC Med • Characteristic of interest • Obtaining relief • H0: π ≥ 0.90 • The claim is correct and any observed sample difference is due to chance • H1: π < 0.90 • The claim is false and the observed sample proportion is unlikely to have come from a population with a π of 0.90 PP 3

  34. Reject p Proportion Problem - OTC Med Sampling Distribution of p • Find critical value • Z.01 = -2.33 • DR: if (Z test statistic ≥ -2.33) Do not reject H0 • What do we observe in the sample? • p = x/n = 160/200 = 0.80 • Create test statistic • Z = (.80 - .90)/ 0212 = -4.71 • Statistical decision • Reject the H0 • Conclusions • The claim is not legitimate p PP 3

  35. Look at t- table PP 3

  36. .80 .90 p - 4.71 0 Z p - Value Approach • How likely are we to observe a sample proportion of 0.80 or one more extreme if the population proportion is 0.90? • Standard normal table • P(0 < Z < 4.71) ≈ 0.50 • There is ≈ .000 in the tail • p - value = .000 • p - value < .01, Reject .000 ≈ 0.50 PP 3

  37. Standard Normal PP 3

  38. Weier’s Z Tables ≈ 0.500 ≈ .000 - 3.10 0 - 3.10 Z Any Z value beyond 3.09 can be viewed as a percentile point that contains roughly 50% of Z values between it and 0 PP 3

  39. Online Homework - Chapter 10 Overview of Hypothesis Testing • CengageNOW fourth assignment PP 3

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