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  1. Hypothesis TestsOne Sample Means

  2. Take a sample & find x. But how do I know if this x is one that I expect to happen or is it one that is unlikelyto happen? How can I tell if they really are underweight? Example 2: A government agency has received numerous complaints that a particular restaurant has been selling underweight hamburgers. The restaurant advertises that it’s patties are “a quarter pound” (4 ounces). A hypothesis test will help me decide!

  3. What are hypothesis tests? Is it one of the sample means that are likely to occur? Calculations that tell us if a value, x, occurs by random chance or not – if it is statistically significant Is it . . . • a random occurrence due to natural variation? • a biased occurrence due to some other reason? Statistically significant means that it is NOT a random chance occurrence! Is it one that isn’t likely to occur?

  4. Nature of hypothesis tests - How does a murder trial work? • First begin by supposing the “effect” is NOT present • Next, see if data provides evidence against the supposition Example: murder trial First - assume that the person is innocent Then – must have sufficient evidence to prove guilty Hmmmmm … Hypothesis tests use the same process!

  5. Notice the steps are the same except we add hypothesis statements – which you will learn today Steps: • Assumptions • Hypothesis statements & define parameters • Calculations • Conclusion, in context

  6. Assumptions for z-test (t-test): Have an SRSof context Distribution is (approximately) normal Given Large sample size Graph data s is known (unknown) YEA – These are the same assumptions as confidence intervals!!

  7. Have an SRS of bottles • Sampling distribution is approximately • normalbecause the boxplot is • symmetrical • s is unknown Example 1:Bottles of a popular cola are supposed to contain 300 mL of cola. There is some variation from bottle to bottle. An inspector, who suspects that the bottler is under-filling, measures the contents of six randomly selected bottles. Are the assumptions met? 299.4 297.7 298.9 300.2 297 301

  8. Writing Hypothesis statements: • Null hypothesis – is the statement being tested; this is a statement of “no effect” or “no difference” • Alternative hypothesis – is the statement that we suspect is true H0: Ha:

  9. The form: Null hypothesis H0: parameter = hypothesized value Alternative hypothesis Ha: parameter > hypothesized value Ha: parameter < hypothesized value Ha: parameter = hypothesized value

  10. Example 2: A government agency has received numerous complaints that a particular restaurant has been selling underweight hamburgers. The restaurant advertises that it’s patties are “a quarter pound” (4 ounces). State the hypotheses : H0: m = 4 Ha: m < 4 Where m is the true mean weight of hamburger patties

  11. Example 3: A car dealer advertises that is new subcompact models get 47 mpg. You suspect the mileage might be overrated. State the hypotheses : H0: m = 47 Ha: m < 47 Where m is the true mean mpg

  12. Example 4: Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40-A fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40. If the mean amperage is lower than 40, customers will complain because the fuses require replacement too often. If the amperage is higher than 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction. State the hypotheses : H0: m = 40 Ha: m = 40 Where m is the true mean amperage of the fuses

  13. Facts to remember about hypotheses: • ALWAYS refer to populations (parameters) • The null hypothesis for the “difference” between populations is usually equal to zero • The null hypothesis for the correlation (rho) of two events is usually equal to zero. H0: mx-y= 0 H0: r= 0

  14. Must use parameter (population) x is a statistics (sample) Activity: For each pair of hypotheses, indicate which are not legitimate & explain why Must be NOT equal! p is the population proportion! Must use same number as H0! r is parameter for population correlation coefficient – but H0MUST be “=“ !

  15. P-values - • Assuming H0 is true,the probability that the test statistic would have a value as extreme or morethan what is actually observed In other words . . . is it far out in the tails of the distribution?

  16. Level of Significance Activity

  17. Level of significance - • Is the amount of evidence necessary before we begin to doubt that the null hypothesis is true • Is the probability that we will reject the null hypothesis, assuming that it is true • Denoted by a • Can be any value • Usual values: 0.1, 0.05, 0.01 • Most common is 0.05

  18. Statistically significant – • The p-value is as small or smaller than the level of significance (a) • If p > a, “fail to reject” the null hypothesis at the a level. • If p <a, “reject” the null hypothesis at the a level.

  19. Facts about p-values: • ALWAYSmake decision about the null hypothesis! • Large p-values show support for the null hypothesis, but never that it is true! • Small p-values show support that the null is not true. • Double the p-value for two-tail (=)tests • Never accept the null hypothesis!

  20. Never“accept” the null hypothesis! Never“accept” the null hypothesis! Never“accept” the null hypothesis!

  21. At an alevel of .05, would you reject or fail to reject H0 for the given p-values? • .03 • .15 • .45 • .023 Reject Fail to reject Fail to reject Reject

  22. Calculating p-values • For z-test statistic – • Use normalcdf(lb,ub) • [using standard normal curve] • For t-test statistic – • Use tcdf(lb, ub, df)

  23. Draw & shade a curve & calculate the p-value: 1) right-tail test t = 1.6; n = 20 2) left-tail test z = -2.4; n = 15 3) two-tail test t = 2.3; n = 25 P-value = .0630 P-value = .0082 P-value = (.0152)2 = .0304

  24. Writing Conclusions: • A statement of the decision being made (reject or fail to reject H0) & why (linkage) • A statement of the results in context. (state in terms of Ha) AND

  25. “Since the p-value < (>) a, I reject (fail to reject) the H0. There is (is not) sufficient evidence to suggest that Ha.” Be sure to write Ha in context (words)!

  26. P-value = tcdf(2.1,10^99,24) =.0232 t=2.1 H0: m = 15 Ha: m > 15 Where m is the true mean concentration of lead in drinking water Example 5: Drinking water is considered unsafe if the mean concentration of lead is greater than 15 ppb (parts per billion). Suppose a community randomly selects of 25 water samples and computes a t-test statistic of 2.1. Assume that lead concentrations are normally distributed. Write the hypotheses, calculate the p-value & write the appropriate conclusion for a = 0.05. Since the p-value < a, I reject H0. There is sufficient evidence to suggest that the mean concentration of lead in drinking water is greater than 15 ppb.

  27. P-value = tcdf(1.9,10^99,11) =.0420 t=1.9 H0: m = 240 Ha: m > 240 Where m is the true mean caloric content of the frozen dinners Example 6: A certain type of frozen dinners states that the dinner contains 240 calories. A random sample of 12 of these frozen dinners was selected from production to see if the caloric content was greater than stated on the box. The t-test statistic was calculated to be 1.9. Assume calories vary normally. Write the hypotheses, calculate the p-value & write the appropriate conclusion for a = 0.05. Since the p-value < a, I reject H0. There is sufficient evidence to suggest that the true mean caloric content of these frozen dinners is greater than 240 calories.

  28. Formulas: s known: m z =

  29. Formulas: s unknown: m t =

  30. Example 7: The Fritzi Cheese Company buys milk from several suppliers as the essential raw material for its cheese. Fritzi suspects that some producers are adding water to their milk to increase their profits. Excess water can be detected by determining the freezing point of milk. The freezing temperature of natural milk varies normally, with a mean of -0.545 degrees and a standard deviation of 0.008. Added water raises the freezing temperature toward 0 degrees, the freezing point of water (in Celsius). The laboratory manager measures the freezing temperature of five randomly selected lots of milk from one producer with a mean of -0.538 degrees. Is there sufficient evidence to suggest that this producer is adding water to his milk?

  31. SRS? Assumptions: Normal? How do you know? • I have an SRS of milk from one producer • The freezing temperature of milk is a normal distribution. (given) Do you know s? • sis known What are your hypothesis statements? Is there a key word? H0: m = -0.545 Ha: m > -0.545 where m is the true mean freezing temperature of milk Plug values into formula. p-value = normalcdf(1.9566,1E99)=.0252 Use normalcdf to calculate p-value. a = .05

  32. Compare your p-value to a & make decision Conclusion: Since p-value < a, I reject the null hypothesis. There is sufficient evidence to suggest that the true mean freezing temperature is greater than -0.545. This suggests that the producer is adding water to the milk. Write conclusion in context in terms of Ha.

  33. Example 8: The Degree of Reading Power (DRP) is a test of the reading ability of children. Here are DRP scores for a random sample of 44 third-grade students in a suburban district: (data on note page) At the a = .1, is there sufficient evidence to suggest that this district’s third graders reading ability is different than the national mean of 34?

  34. H0: m = 34 where m is the true mean reading Ha: m = 34 ability of the district’s third-graders SRS? • I have an SRS of third-graders Normal? How do you know? • Since the sample size is large, the sampling distribution is approximately normally distributed • OR • Since the histogram is unimodal with no outliers, the sampling distribution is approximately normally distributed Do you know s? What are your hypothesis statements? Is there a key word? • sis unknown Plug values into formula. p-value = tcdf(.6467,1E99,43)=.2606(2)=.5212 Use tcdf to calculate p-value. a = .1

  35. Compare your p-value to a & make decision Conclusion: Since p-value > a, I fail to reject the null hypothesis. There is not sufficient evidence to suggest that the true mean reading ability of the district’s third-graders is different than the national mean of 34. Write conclusion in context in terms of Ha.

  36. Example 9: The Wall Street Journal (January 27, 1994) reported that based on sales in a chain of Midwestern grocery stores, President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Does this indicate that the sales of the cookies is different from the earlier figure?

  37. Assume: • Have an SRS of weeks • Distribution of sales is approximately normal due to large sample size • s unknown • H0: m = 1323 where m is the true mean cookie sales • Ha: m≠ 1323 per week • Since p-value < a of 0.05, I reject the null hypothesis. There is sufficient to suggest that the sales of cookies are different from the earlier figure.

  38. Example 9: President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Compute a 95% confidence interval for the mean weekly sales rate. CI = ($1105.30, $1310.70) Based on this interval, is the mean weekly sales rate statistically different from the reported $1323?

  39. In a one-sided test, all of a (2%) goes into that tail (lower tail). What do you notice about the decision from the confidence interval & the hypothesis test? Remember your, p-value = .01475 At a = .02, we would reject H0. What decision would you make on Example 10 if a = .01? What confidence level would be correct to use? Does that confidence interval provide the same decision? If Ha: m < 1323, what decision would the hypothesis test give at a = .02? Now, what confidence level is appropriate for this alternative hypothesis? A 96% CI = ($1100, $1316). Since $1323 is not in the interval, we would reject H0. You would fail to reject H0 since the p-value > a. You should use a 99% confidence level for a two-sided hypothesis test at a = .01. The 98% CI = ($1084.40, $1331.60) - Since $1323 is in the interval, we would fail to reject H0. Why are we getting different answers? Tail probabilities between the significant level (a) and the confidence level MUST match!) In a CI, the tails have equal area – so there should also be 2% in the upper tail CI = ($1068.6 , $1346.40) - Since $1323 is in this interval we would fail to reject H0. a = .02 .96 .02 That leaves 96% in the middle & that should be your confidence level

  40. Matched Pairs Test A special type of t-inference

  41. Pair individuals by certain characteristics Randomly select treatment for individual A Individual B is assigned to other treatment Assignment of B is dependent on assignment of A Individual persons or items receive both treatments Order of treatments are randomly assigned or before & after measurements are taken The two measures are dependent on the individual Matched Pairs – two forms

  42. 1)A college wants to see if there’s a difference in time it took last year’s class to find a job after graduation and the time it took the class from five years ago to find work after graduation. Researchers take a random sample from both classes and measure the number of days between graduation and first day of employment Is this an example of matched pairs? No, there is no pairing of individuals, you have two independent samples

  43. 2) In a taste test, a researcher asks people in a random sample to taste a certain brand of spring water and rate it. Another random sample of people is asked to taste a different brand of water and rate it. The researcher wants to compare these samples Is this an example of matched pairs? No, there is no pairing of individuals, you have two independent samples – If you would have the same people taste both brands in random order, then it would bean example of matched pairs.

  44. 3) A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to a random sample, company researchers take a weight measurement on each person. After a month of using the drug, each person’s weight is measured again. Is this an example of matched pairs? Yes, you have two measurements that are dependent on each individual.

  45. A whale-watching company noticed that many customers wanted to know whether it was better to book an excursion in the morning or the afternoon. To test this question, the company collected the following data on 15 randomly selected days over the past month. (Note: days were not consecutive.) You may subtract either way – just be careful when writing Ha Since you have two values for each day, they are dependenton the day – making this data matched pairs First, you must find the differences for each day.

  46. I subtracted: Morning – afternoon You could subtract the other way! • Assumptions: • Have an SRS of days for whale-watching • s unknown • Since the normal probability plot is approximately linear, the distribution of difference is approximately normal. You need to state assumptions using the differences! Notice the granularity in this plot, it is still displays a nice linear relationship!

  47. Is there sufficient evidence that more whales are sighted in the afternoon? Be careful writing your Ha! Think about how you subtracted: M-A If afternoon is more should the differences be + or -? Don’t look at numbers!!!! If you subtract afternoon – morning; then Ha: mD>0 H0: mD = 0 Ha: mD < 0 Where mD is the true mean difference in whale sightings from morning minus afternoon Notice we used mD for differences & it equals 0 since the null should be that there is NO difference.

  48. finishing the hypothesis test: Since p-value > a, I fail to reject H0. There is insufficient evidence to suggest that more whales are sighted in the afternoon than in the morning. In your calculator, perform a t-test using the differences (L3) Notice that if you subtracted A-M, then your test statistic t = + .945, but p-value would be the same