"Hypothesis Tests One Sample Means"
Hypothesis Tests One Sample Means 2: can I tell if they really ExampleHowA government agency are underweight? has received numerous complaints that a particular Hypothesis test restaurant has been selling will help me decide! a sample & find The underweight hamburgers. x. Take restaurant advertises that it’s patties are “a quarter pound” (4 But how do I know if this x is one ounces). I expect to happen or is it one that that is unlikely to happen? What are hypothesis tests? Calculations that tell us if a value occurs by random chance or not – if it is statistically significant Is it . . . – a random occurrence due to variation? – a biased occurrence due to some other reason? Nature of hypothesis tests - How does a murder trial work? • First begin by supposing the First assume present “effect”- is NOTthat the • Next, see if data provides person is innocent Then – must have the evidence againstsufficient evidence suppositionto prove guilty Hmmmmm … Example: murder Hypothesis tests use trial the same process! Notice the steps are the Steps: same except we add hypothesis statements – which you will learn today 1) Assumptions 2) Hypothesis statements & define parameters 3) Calculations 4) Conclusion, in context Assumptions for z-test (t-test): YEA – These are the same • Have an SRS of context assumptions as confidence • intervals!! Distribution is (approximately) normal – Given – Large sample size – Graph data • s is known (unknown) Example 1: Bottles of a popular cola •Have an SRS to contain are supposed of bottles 300 mL of cola. There is some variation from •Sampling distribution is approximately bottle to bottle. An inspector, who normal because the boxplot is symmetrical suspects that the bottler is under- • s is unknown filling, measures the contents of six randomly selected bottles. Are the assumptions met? 299.4 297.7 298.9 300.2 297 301 Writing Hypothesis statements: • Null hypothesis – is the statement being tested; this is a statement of “no effect” or “no difference” H0: • Alternative hypothesis – is the statement that we suspect is true Ha: The form: Null hypothesis H0: parameter = hypothesized value Alternative hypothesis Ha: parameter > hypothesized value Ha: parameter < hypothesized value Ha: parameter = hypothesized value 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 Where m is the true mean weight of Ha: m < 4 hamburger patties 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 Where m is the true mean mpg Ha: m < 47 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 Where m is the true mean amperage of Ha: m = 40 the fuses Facts to remember about hypotheses: • ALWAYS refer to populations (parameters) • The null hypothesis for the “difference” between populations is usually equal to zero H0: mx-y= 0 • The null hypothesis for the correlation (rho) of two events is usually equal to zero. H0: r= 0 Activity: For each pair of Must use indicate (population) x Must be NOT are hypotheses,parameter whichequal!not is a explain (sample) legitimate & statisticswhy a) H 0 : m 15 ; H a : m 15 is the population Ha : x b) H 0 : x 123;proportion!123 Must use same : . as H0 r is number1; H : .1 c) H 0parameter fora! population correlation coefficient – but H0 d) H 0 : m .4;be “=“ !m .6 MUST H a : e) H 0 : r 0 ; H a : r 0 P-values - • The probability that the test statistic would have a value as extreme or more than what is actually observed In other words . . . is it far out in the tails of the distribution? Level of Significance Activity 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 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. Facts about p-values: • ALWAYS make 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! Never “accept” the null hypothesis! Never “accept” the null hypothesis! Never “accept” the null hypothesis! At an a level of .05, would you reject or fail to reject H0 for the given p-values? a) .03 Reject b) .15 Fail to reject c) .45 Fail to reject d) .023 Reject Calculating p-values • For z-test statistic – – Use normalcdf(lb,ub) – [using standard normal curve] • For t-test statistic – – Use tcdf(lb, ub, df) 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 Writing Conclusions: 1) A statement of the decision being made (reject or fail to reject H0) & why (linkage) AND 2) A statement of the results in context. (state in terms of Ha) “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)! Example 5: Drinking water is considered unsafe if the mean H0: m = 15 concentration of lead is 15 ppb (parts Ha: billion) per m > 15 or greater. Suppose a t=2.1 Where m is the true mean concentration community randomly selects 25 water of lead in drinking water a t-test samples and computesreject H0. There is Since the p-value < a, I sufficient evidence to suggest lead P-value of 2.1. Assume that statistic = tcdf(2.1,10^99,24) that the mean concentration normally concentrations are of lead in drinking =.0232 water is greater than 15 hypotheses, distributed. Write the ppb. calculate the p-value & write the appropriate conclusion for a = 0.05. Example 6: A certain type of frozen dinners states that the dinner contains 240 calories. A random H0: m = 240 sample: of> 12 of these frozen dinners Ha m 240 Where m is thet=1.9 mean caloric see was selected fromtrue production to Since the p-value < a, I reject H0. There is content of the frozen dinners if the caloric content was greater sufficient evidence to suggest that the than stated on the box. The t-test P-value = caloric content of true meantcdf(1.9,10^99,11) these frozen =.0420 was calculated to be 1.9. statistic is greater than 240 calories. dinners Assume calories vary normally. Write the hypotheses, calculate the p-value & write the appropriate conclusion for a = 0.05. Formulas: s known: statistic - parameter test statistic standard deviation of statistic z= x m σ n Formulas: s unknown: statistic - parameter test statistic standard deviation of statistic t= x m s n 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? SRS? Assumptions: Normal? •I have an SRS of milk from one producer How do you •The freezing temperature of milk is a normal distribution. know? (given) • s is known Do you H0: m = -0.545 know s? What are your hypothesis Ha: m > -0.545 statements? Is where m is the true mean freezing temperature of milk word? there a key .538 .545 z 1.9566 Plug values .008 into formula. 5 p-value = normalcdf(1.9566,1E99)=.0252 Use normalcdf to a = .05 calculate p-value. Conclusion: Compare your p-value to a & make decision 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. 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? SRS? • I have an SRS of third-graders Normal? •Since the sample size is large, the sampling distribution is approximately normally distributed How do you know? OR Do you know s? •Since the histogram is unimodal with no outliers, the are your What sampling distribution is approximately normally distributed hypothesis • s is unknown statements? Is H0: m = 34 where m is the true mean reading there a key word? Ha: m = 34 ability of the district’s third-graders 35.091 34 Plug values t .6467 11.189 into formula. 44 p-value = tcdf(.6467,1E99,43)=.2606(2)=.5212 Use tcdf to calculate a = .1 p-value. Conclusion: Compare your p-value to a & make decision 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. 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? 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 per Ha: m ≠ 1323 week 1208 1323 t 2.29 p value .0295 275 30 Since p-value < a of 0.05, I reject the null hypothesis. There is sufficient evidence to suggest that the sales of cookies are different from the earlier figure. 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? What do you notice about the decision from the confidence interval & the hypothesis test? What decision would you make on Example 9 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 = .05? a = .01? Now, what confidence levels are appropriate for this alternative hypothesis? Matched Pairs Test A special type of t- inference Matched Pairs – two forms • Pair individuals by • Individual persons certain characteristics or items receive both • Randomly select treatments treatment for • Order of treatments individual A are randomly • Individual B is assigned or before & assigned to other after measurements treatment are taken • Assignment of B is • The two measures dependent on are dependent on assignment of A the individual Is this an example of matched pairs? 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 No, there is no pairing of individuals, you have two independent samples Is this an example of matched pairs? 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 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 be an example of matched pairs. Is this an example of matched pairs? 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. Yes, you have two measurements that are dependent on each individual. 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 You may subtract either way – data careful randomly selected days over followingjust be on 15 when writing Ha the past month. (Note: days were not consecutive.) Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Morning 8 9 7 9 10 13 10 8 2 5 7 7 6 8 7 After- noon 8 10 9 8 9 11 8 10 4 7 8 9 6 6 9 Since you have two values for First, you must find the each day, they are dependent on differences for each the day – making this data day. matched pairs Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Morning 8 9 7 9 10 13 10 8 2 5 7 7 6 8 7 After- noon 8 10 9 8 9 11 8 10 4 7 8 9 6 6 9 I subtracted: Differenc es 0 -1 -2 1 1 2 2 – -2 -2 Morning -2afternoon -1 -2 0 2 -2 Assumptions: You could subtract the other way! • Have an SRS of days for whale-watching You need to state assumptions using the • s unknown differences! •Since the normal probability plot is approximately linear, the distribution of difference is approximately normal. Notice the granularity in this plot, it is still displays a nice linear relationship! Differenc es 0 -1 -2 1 1 2 2 -2 -2 -2 -1 -2 0 2 -2 Is there sufficient evidence that more whales are sighted in the afternoon? Be careful writing your Ha! H0: mD = 0 If you subtract afternoonhow you Think about – morning; subtracted: M-A then Ha: m >0 Notice we used mD for differences the If afternoon isDmore should Ha: mD < 0 differences should -? & it equals 0 since the null be + or be that there is NO difference. Don’t look at numbers!!!! Where mD is the true mean difference in whale sightings from morning minus afternoon Differenc es 0 -1 -2 1 1 2 2 -2 -2 -2 -1 -2 0 2 -2 finishing the hypothesis test: In your calculator, xμ .4 0 perform a t-test t .945 Notice that if you using the differences s 1.639 subtracted A-M, n 15 (L3) then your test p .1803 statistic t = + .945, but p- df 14 α .05 value would be the same 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.