Statistics for Business and Economics: Confidence Intervals for Means

Transcription

1 Statistics for Business and Economics: Confidence Intervals for Means STT 315: Section 201 Instructor: Abdhi Sarkar Acknowledgement: I d like to thank Dr. AshokeSinha for allowing me to use and edit the slides.

2 Set-up Suppose we take a random sample of size n from a population with mean and standard deviation. The sample mean will serve the purpose of point estimator of population mean. Goal: To construct a %C.I. for. However the procedure will depend on whether the sample size n is large enough or not, we know the value of or not. 2

3 Large sample C.I. s of 3

4 Reminder: Sampling distribution of Suppose we take a random sample from a population with mean, and standard deviation. In that case, the sample mean has the following properties: = =. =. Furthermore, for large sample size ( 30) ~,, approximately. 4

5 Building a C.I. for If ~(0,1)then / is such a number that > =. Thus P < < = 1. Since ~, approximately, we have $% ~ 0,1 approximately. So working backward we find that there is roughly 1 probability that the interval, + contain. will 5

6 100 1 %C.I. for If sample size is largethen the % approximate C.I. for is: (, if std. dev. ()is known,, if std. dev. is unknown, where is the sample mean, and )is the sample standard deviation. If 30, we can consider the sample is large enough. If sample is not large enough, we need to assume that the population is normally distributed. We shall use TI83/84 to compute C.I. s for. 6

7 Example A sample of 82 MSU undergraduates, the mean number of Facebook friends was friends with standard deviation of friends. Use this information to make a 95% confidence interval for the average number of Facebook friends MSU undergraduates have. Press [STAT]. Select [TESTS]. Choose 7: ZInterval. Select with arrow keys Stats Input the following: : : n : 82 C-Level: 95 Choose Calculate and press [ENTER]. Answer: 95% C.I. for µ is (520.19, ). 7

8 C.I. s of for normal populations 8

9 Reminder: Sampling distribution of Suppose we take a random sample from a population normally distributed with mean,and standard deviation. In that case, the sample mean has the following properties: = =. =. Furthermore, if the population is normally distributed then ~,,. 9

10 100 1 %C.I. for [known ] If the sample is from normally distributedpopulation with known std. dev., then the %C.I. for is:, where is the sample mean. Use ZInterval from TI 83/84 to compute C.I. for [known ]. The margin of error: M.E.=. The width of the C.I. is 2 = 2-.. To find use: =./012 1,0,1. 10

11 100 1 %C.I. for [known ] Larger the std. dev.,larger the M.E. Larger the confidence level, larger the M.E. Larger the sample size, smaller the M.E. Given the confidence level and std. dev., one can find the optimal sample size for a particular margin of error using the formula: = Always round-up for the optimal sample size. 11

12 Example The number of bolts produced each hour from a particular machine is normally distributed with a standard deviation of 7.4. For a random sample of 15 hours, the average number of bolts produced was Find a 98% confidence interval for the population mean number of bolts produced per hour. Press [STAT]. Select [TESTS]. Choose 7: ZInterval. Select with arrow keys Stats Input the following: : 7.4 : n : 15 C-Level: 98 Choose Calculate and press [ENTER]. Answer: 98% C.I. for µ is (582.86, ). 12

13 Example The number of bolts produced each hour from a particular machine is normally distributed with a standard deviation of 7.4. For a random sample of 15 hours, the average number of bolts produced was Find a 98% confidence interval for the population mean number of bolts produced per hour. We found 98% C.I. for µ is (582.86, ). Width = =8.88. So M.E = Width/2 = Suppose we want the margin of error for 98% confidence interval for the population mean number of bolts produced per hour to be 3.5. What is the optimal sample size? We shall use = For 98% C.I., = So = 9.9: =./ ,0,1 = =.B = So optimal sample size is

14 100 1 %C.I. for [unknown and n<30 ] However the formula of the previous C.I. of cannot be used if the std. dev. is unknown. In such case, one should substitute by sample standard deviation ). However, unlike the large sample we can no longer use - distribution (i.e. (0,1) distribution). In that case, student s D-distribution comes to rescue. D-distributions are all symmetric continuous distributions centered around 0. A degree of freedom (EF) is attached to each D-distrn. For our problem, EF = 1. 14

15 The concept of D ;HI If J~D HI then D5 ;HI is such a number that 6 J > D ;HI =. 2 Thus P D ;HI <J <D ;HI =1. 15

16 100 1 %C.I. for [unknown ] If the sample is from normally distributedpopulation but the std. dev. is unknown, then the % C.I. for is: D ;$: ), where is the sample mean, and )is the sample standard deviation. Here the margin of error is D ;$: (. The width of the C.I. is 2D ;$: ( = 2-.. Use TInterval from TI 83/84 to compute C.I. for [unknown ]. 16

17 Example The Daytona Beach Tourism Commission is interested in the average amount of money a typical college student spends per day during spring break. They survey 25 students and find that the mean spending is $63.57 with a standard deviation of $ Develop a 97% confidence interval for the population mean daily spending. Press [STAT]. Select [TESTS]. Choose 8: TInterval. Select with arrow keys Stats Input the following: : Sx: n : 25 C-Level: 97 Choose Calculate and press [ENTER]. Answer: 97% C.I. for µ is (55.58, 71.56). 17