Statistics for Business and Economics: Confidence Intervals for Proportions

Transcription

1 Statistics for Business and Economics: Confidence Intervals for Proportions STT 315: Section 107 Acknowledgement: I d like to thank Dr. Ashoke Sinha for allowing me to use and edit the slides.

2 Statistical Inference Inference means that we are making a conclusion about the population parameter based on the statistic we calculated from a sample. Conclusions made using statistical inference are probabilistic in nature. We may not be able to say for sure, but with certain confidence. There are two types of inference: Confidence Intervals, Hypothesis Tests. 2

3 Students will be able to: Goal Construct a confidence interval for a proportion. Interpret a confidence interval for a proportion. Check conditions for the use of inference about a population proportion Independence (or sample less than 10% of population), Sample size large enough (successes and failures each greater than 10). Explain the relationship between the margin of error, sample size, and level of certainty. 3

4 Estimating Smokers Suppose I want to estimate the percent of MSU undergraduate students who smoke. A random sample of 99 undergraduate students were selected and 17 of them smoked tobacco last week. I want to make a 95% confidence interval for the proportion of MSU undergraduates based on this information. 4

5 Are the conditions met? Firstly it is a random sample. Though the sample is without replacement, but it satisfies 10% condition as there are more than 1000 undergraduate students in MSU. Also both number of smokers (17) and non-smokers (82) are larger than 10, the sample can be considered to be large enough. 5

6 Use the results to make a CI The sampling distribution results guarantees us that the sample proportions will be roughly normally distributed around the population proportion. So 95% of samples should fall within two standard deviations of the population proportion. But we don t know the population proportion (that s what we are trying to estimate)! So we cannot get ˆp. Therefore we need to use the sample proportion and work backward from there. 6

7 Construction of C.I. In our sample, 17 out of 99 students smoked tobacco in the last week, or 17.2%. 17.2% is a statistic (or a point estimate). We will use 17.2% to make an interval estimate for the value of the parameter. To make a 95% confidence interval we must create an interval that is 2 standard deviations long, above and below the statistic. One standard deviation is.172(.828) So 2 standard deviations is 2(.0379) =.0758 (and 7.58% is the margin of error). 7

8 So a 95% confidence interval has endpoints at = , and = And we write: We are 95% confident that between 9.62% and 24.78% of MSU undergraduates smoke tobacco. If we want to make a 68% confidence interval, we only have to extend the interval one standard deviation from the statistic in each direction: So a 68% confidence interval has endpoints at = , and = and we write: We are 68% confident that between 13.4% and 21.0% of MSU undergraduates smoke tobacco. 8

9 Our 95% CI for smokers was 9.62% to 24.78%. This means that (find the correct one): a) 95% of random samples of MSU undergraduates will have between 9.62% and 24.78% smokers. b) Between 9.62% and 24.78% of MSU undergraduates smoke. c) 95% of MSU undergraduates smoke between 9.62% and 24.78% of the time. d) We are 95% sure that between 9.62% and 24.78% of MSU undergraduates smoke. 9

10 Standard Error (S.E.) If subjects are independent, and if the sample size is large enough, then the sample proportions are approximately normally distributed with mean p, and standard p(1 p) deviation. pˆ n i.e., pˆ ~ N( p, ˆp ) approximately. But in an estimation problem, p is unknown. So we replace population proportion (p) by the sample proportion ( ) in its formula and get standard error of sample proportion S. E.( pˆ) pˆ pˆ(1 pˆ) pq ˆ ˆ, where qˆ 1 pˆ. n n 10

11 Confidence Interval (C.I.) and Margin of Error (M.E.) Since for large n, the sample proportion ( pˆ ) is approximately normal, we can conclude (using empirical rule) that within a margin of error of 1 S.E. we are about 68% sure the population proportion (p) lies. within a margin of error of 2 S.E. we are about 95% sure the population proportion (p) lies. So confidence interval for p is pˆ M. E. Obviously, more the confidence you require, larger the margin of error. 11

12 Find the exact area between -2 and 2 standard deviations from the mean on a normal curve using your calculator. Hint: normalcdf(-2,2,0,1) = = 95.4%. So this is not exactly 95%, but slightly more. On the other hand the exact area between and 1.96 standard deviations from the mean is normalcdf(-1.96,1.96,0,1) = 0.95 = 95%. Using 1.96, we get the 95% C.I. for p to be: (0.097, 0.246). Is it 1.96 or 2 for 95% C.I.? Note: calculator uses But how to use calculator to construct C.I.? 12

13 Formula: α % C.I. for p The formula for α % C.I. for p is given by p z Τ α 2 p 1 p n = p z ατ2 SE p, where z ατ2 is such a number that P൫Z > z ατ2 ൯ = α, where Z is a standard normal 2 variable. However, one can use TI 83/84 to compute C.I. s for p. 13

14 C.I. with TI 83/84 Plus Want to make a 85% confidence interval for smokers among MSU undergraduates. In a random sample of 99 MSU undergraduates 17 smoked tobacco last week. Press [STAT]. Select [TESTS]. Choose A: 1-PropZInt. Input the following: o x: 17 o n: 99 o C-Level: 85 Choose Calculate and press [ENTER]. Answer: 85% C.I. for p is (0.117, 0.226). 14

15 How would I a confidence interval a proportion using a calculator? The sample input shows finding a 99% confidence interval with a sample size of 4040 people and 2048 smokers. We would interpret the sample output as: We are 99% confident that between 48.7% and 52.7% of the population smokes. Sample Input Sample Output Note: This example wasn t actually about smoking. 15

16 Width of a C.I. Since the formula of confidence interval for p is the width of the C.I. is 2 M.E. So if we know the width of C.I., we can compute the M.E. by halving the width. Example: Given a 90% C.I. for p is (0.23, 0.37), find the values of (a) sample proportion and (b) margin of error of the 90% C.I. Solution: Since the width = ( ) = 0.14, and so the margin of error of 90% C.I. for p is 0.14/2 = Moreover and so pˆ pˆ M. E. pˆ M. E

17 Smoker example We found that 17.2% of a sample of 99 MSU undergraduates had smoked in the past week. We used this to find a 95% confidence interval for the proportion of MSU undergraduates who smoke. The endpoints of our 95% confidence interval is (0.096, 0.248). The width of 95% C.I. is ( ) = 0.15, and so the margin of error is (0.15/2) = If we want to reduce the margin of error while keeping the confidence level the same, we could increase the sample size. 17

18 M.E. and sample size If we wanted to reduce the margin of error to 4%, minimum how many undergrads would we have to survey? The formula is: n 2 ( z / 2) pq. 2 ( M. E.) But what p to use (remember q = 1-p)? Two cases: No information about p is given. In that case use p = 0.5. In our exercise, if nothing about p is known: n n 2 ( M. E.) So we would need 601 subjects (0.04) pq (.172)(.828) ( M. E.) (.04) If some information about p is known, use that information. If we use the information of sample: p = 0.172, q = = So we would need 342 subjects. 18

19 Summary Larger sample size makes smaller margin of error. Larger confidence makes larger margin of error. The level of confidence is the proportion of intervals that will contain the value of the population parameter. As long as the conditions are met, the process of confidence interval works. 19

20 Statistics for Business and Economics: Confidence Intervals for Means

21 Set-up Suppose we take a random sample of size n from a population with mean μ and standard deviation σ. The sample mean xҧ will serve the purpose of point estimator of population mean μ. Goal: To construct a α % C.I. for μ. However the procedure will depend on whether the sample size n is large enough or not, we know the value of σ or not. 21

22 Large sample C.I. s of μ 22

23 Reminder: Sampling distribution of xҧ Suppose we take a random sample from a population with mean μ, and standard deviation σ. In that case, the sample mean xҧ has the following properties: μ x ҧ = E σ x ҧ = σ. n x ҧ = μ. Furthermore, for large sample size (n 30) x~n ҧ μ x ҧ, σ xҧ N μ, σ n approximately. 23

24 Building a C.I. for μ If Z~N(0,1) then z α/2 is such a number that P Z > z Τ α 2 = α 2. Thus P൫ z ατ2 < Z < ൯ = 1 α. z ατ2 Since x~n ҧ μ, σ n approximately. ҧ approximately, we have Z x μ ~N 0,1 στ n So working backward we find that there is roughly 1 α probability that the interval contain μ. xҧ z ατ2 σn, x ҧ + z ατ2 σ n will 24

25 100 1 α % C.I. for μ If sample size is large then the α % approximate C.I. for μ is: xҧ z ατ2 σn xҧ z ατ2 sn, if std. dev. (σ) is known,, if std. dev. is unknown, where xҧ is the sample mean, and s is the sample standard deviation. If n 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 μ. 25

26 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: σ: x ҧ : n : 82 C-Level: 95 Choose Calculate and press [ENTER]. Answer: 95% C.I. for µ is (520.19, ). 26

27 C.I. s of μ for normal populations 27

28 Reminder: Sampling distribution of xҧ Suppose we take a random sample from a population normally distributed with mean μ, and standard deviation σ. In that case, the sample mean xҧ has the following properties: μ x ҧ = E x ҧ = μ. σ x ҧ = σ. n Furthermore, if the population is normally distributed then σ x~n ҧ μ x ҧ, σ xҧ N μ, n. 28

29 100 1 α % C.I. for μ [known σ] If the sample is from normally distributed population with known std. dev. σ, then the α % C.I. for μ is: xҧ z ατ2 σn, where xҧ is the sample mean. Use ZInterval from TI 83/84 to compute C.I. for μ [known σ]. The margin of error: M. E. = z ατ2 σn. The width of the C.I. is 2z Τ α 2 σn = 2M. E. To find z Τ α 2 use: z Τ α 2 = invnorm 1 α 2, 0,1. 29

30 100 1 α % C.I. for μ [known σ] Larger the std. dev. σ, larger the M.E. Larger the confidence level, larger the M.E. Larger the sample size n, 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: n = z 2 ατ2σ. m. e. Always round-up for the optimal sample size n. 30

31 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: σ: 7.4 x ҧ : n : 15 C-Level: 98 Choose Calculate and press [ENTER]. Answer: 98% C.I. for µ is (582.86, ). 31

32 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 n = z 2 ατ2σ. For 98% C.I., α = m.e. So z ατ2 = z 0.01 = invnorm 0.99,0,1 = n = = So optimal sample size is

33 100 1 α % C.I. for μ [unknown σ and n<30 ] However the formula of the previous C.I. of μ cannot be used if the std. dev. σ is unknown. In such case, one should substitute σ by sample standard deviation s. However, unlike the large sample we can no longer use zdistribution (i.e. N(0,1) distribution). In that case, student s t-distribution comes to rescue. t-distributions are all symmetric continuous distributions centered around 0. A degree of freedom (df) is attached to each t-distrn. For our problem, df = n 1. 33

34 The concept of t Τ α 2;df If T~t df then tα ;df is such a number that 2 P T > t ατ2;df = α 2. Thus P t Τ α 2;df < T < t Τ α 2;df = 1 α. 34

35 100 1 α % C.I. for μ [unknown σ] If the sample is from normally distributed population but the std. dev. is unknown, then the α % C.I. for μ is: xҧ t ατ2;n 1 sn, where xҧ is the sample mean, and s is the sample standard deviation. Here the margin of error is t ατ2;n 1 sn. The width of the C.I. is 2t ατ2;n 1 = 2M. E. sn Use TInterval from TI 83/84 to compute C.I. for μ [unknown σ]. 35

36 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: x ҧ : Sx: n : 25 C-Level: 97 Choose Calculate and press [ENTER]. Answer: 97% C.I. for µ is (55.58, 71.56). 36