1. Chapter 11.1 Inference for the Mean of a Population.

2. Example 1: One concern employers have about the use of technology is the amount of time that employees spend each day making personal use of company technology, such as phone, e-mail, internet, and games. The Associated Press reports that, on average, workers spend 72 minutes a day on such personal technology uses. A CEO of a large company wants to know if the employees of her company are comparable to this survey. In a random sample of 10 employees, with the guarantee of anonymity, each reported their daily personal computer use. The times are recorded at right. EmployeeTime 166 270 375 488 569 671 7 863 989 1086 What is different about this problem? When the standard deviation of a statistic is estimated from the data, the result is called the standard error of the statistic, and is given by s/√n. When we use this estimator, the statistic that results does not have a normal distribution, instead it has a new distribution, called the t-distribution. Does the data provide evidence that the mean for this company is greater than 72 minutes?

3. Time for some Nspiration!

4. One-Sample z-statistic  known: z = 

5. One-sample t-statistic:  unknown: t = 

6. The variability of the t-statistic is controlled by the Sample Size. The number of degrees of freeom is equal to n-1.

7. ASSUMING NORMALITY? 1.SRS is extremely important. 2.Check for skewness. 3.Check for outliers. 4.If necessary, make a cautionary statement. 5.In Real-Life, statisticians and researchers try very hard to avoid small samples. Use a Box and Whisker to check.

8. Example 2: 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: 40263914421825 43462719471926 35341544403831 46522535353329 34414928524735 48223341512714 5445 At the  =.1, is there sufficient evidence to suggest that this district’s third graders reading ability is different than the national mean of 34?

9. I have an SRS of third-graders 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  is unknown SRS? Normal? How do you know? Do you know  ? What are your hypothesis statements? Is there a key word? Plug values into formula. p-value = tcdf(.6467,1E99,43)=.2606(2)=.5212 Use tcdf to calculate p-value.  =.1 H 0 :  = 34where  is the true mean reading H a :  = 34 ability of the district’s third-graders Name the Test!! One Sample t-test for mean

10. Conclusion: Compare your p-value to  & make decision Since p-value > , I fail to reject the null hypothesis. Write conclusion in context in terms of H a. 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.

11. Back to Example 1. The times are recorded below. Employee1 2 3 4 5 6 7 8 9 10 Time66 70 75 88 69 71 71 63 89 86 Does this data provide evidence that the mean for this company is greater than 72 minutes?

12. I have an SRS of employees Since the histogram has no outliers and is roughly symmetric, the sampling distribution is approximately normally distributed  is unknown, therefore we are using a 1 sample t-test SRS? Normal? How do you know? Do you know  ? Plug values into formula. p-value = tcdf(.937,1E99,9)=.1866(2)=.3732 Use tcdf to calculate p-value. H 0 :  = 72where  is the true # of min spent on PT H a :  = 72 time spent by this company’s employees What are your hypothesis statements? Is there a key word?

13. Conclusion: Compare your p-value to  & make decision Since p-value > , I fail to reject the null hypothesis that this company’s employees spend 72 minutes on average on Personal Technology uses. Write conclusion in context in terms of H a. There is not sufficient evidence to suggest that the true amount of time spent on personal technology use by employees of this company is more than the national mean of 72 min.

14. Now for the fun calculator stuff!

15. Example 3: 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?

16. Assume: Have an SRS of weeks Distribution of sales is approximately normal due to large sample size s unknown H 0 :  = 1323 where  is the true mean cookie sales H a :  ≠ 1323 per week Since p-value <  of 0.05, I reject the null hypothesis. There is sufficient to suggest that the sales of cookies are different from the earlier figure. Name the Test!! One Sample t-test for mean

17. Example 3: 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?

18. What do you notice about the decision from the confidence interval & the hypothesis test? What decision would you make on Example 3 if  =.01? What confidence level would be correct to use? Does that confidence interval provide the same decision? If H a :  < 1323, what decision would the hypothesis test give at  =.02? Now, what confidence level is appropriate for this alternative hypothesis? You should use a 99% confidence level for a two-sided hypothesis test at  =.01. You would fail to reject H 0 since the p-value > . CI = ($1068.6, $1346.40) - Since $1323 is in this interval we would fail to reject H 0. Remember your, p-value =.01475 At  =.02, we would reject H 0. The 98% CI = ($1084.40, $1331.60) - Since $1323 is in the interval, we would fail to reject H 0. Why are we getting different answers? In a one-sided test, all of  goes into that tail (lower tail).  =.02 In a CI, the tails have equal area – so there should also be 2% in the upper tail 96% confidence level That leaves 96% in the middle & that should be your confidence level. 02.96 A 96% CI = ($1100, $1316). Since $1323 is not in the interval, we would reject H 0. Tail probabilities between the significant level (  ) and the confidence level MUST match!)

19. Ex4: The times of first sprinkler activation (seconds) for a series of fire-prevention sprinklers were as follows: 27 4122272335303324 27282224 Construct a 95% confidence interval for the mean activation time for the sprinklers.

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

21. Matched Pairs – two forms 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

22. 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

23. 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.

24. 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.

25. 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.) Day123456789101112131415 Morning 897910131082577687 After- noon 810989118104789669 First, you must find the differences for each day. Since you have two values for each day, they are dependent on the day – making this data matched pairs You may subtract either way – just be careful when writing H a

26. Day 123456789101112131415 Morning 897910131082577687 After- noon 810989118104789669 Differenc es 0-21122 -202 Assumptions: Have an SRS of days for whale-watching  unknown Since the boxplot doesn’t show any outliers, we can assume the distribution is approximately normal. I subtracted: Morning – afternoon You could subtract the other way! You need to state assumptions using the differences! Notice the skewness of the boxplot, however, with no outliers, we can still assume normality!

27. Differences 0-21122 -202 Is there sufficient evidence that more whales are sighted in the afternoon? Be careful writing your H a ! Think about how you subtracted: M-A If afternoon is more should the differences be + or -? Don’t look at numbers!!!! H 0 :  D = 0 H a :  D < 0 Where  D is the true mean difference in whale sightings from morning minus afternoon Notice we used  D for differences & it equals 0 since the null should be that there is NO difference. If you subtract afternoon – morning; then H a :  D >0

28. finishing the hypothesis test: Since p-value > , I fail to reject H 0. There is insufficient evidence to suggest that more whales are sighted in the afternoon than in the morning. Notice that if you subtracted A-M, then your test statistic t = +.945, but p- value would be the same In your calculator, perform a t-test using the differences (L3) Differences 0-21122 -202

29. PlayerBeforeAfter 11318 22037 31740 41335 51330 61620 71533 81619 Ex: The effect of exercise on the amount of lactic acid in the blood was examined in journal Research Quarterly for Exercise and Sport. Eight males were selected at random from those attending a week-long training camp. Blood lactate levels were measured before and after playing 3 games of racquetball, as shown in the table. What is the parameter of interest in this problem? Construct a 95% confidence interval for the mean change in blood lactate level.

30. Based on the data, would you conclude that there is a significant difference, at the 5% level, that the mean difference in blood lactate level was over 10 points? PlayerBeforeAfter 11318 22037 31740 41335 51330 61620 71533 81619