# INTRODUCTION DERIVATION NOTATIONS AND GENERAL FORM 1 SAMPLE Z-INTERVAL FOR µ 1 SAMPLE T-INTERVAL FOR µ 2 INDEPENDENT SAMPLE T-INTERVAL FOR µ 2 DEPENDENT.

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INTRODUCTION DERIVATION NOTATIONS AND GENERAL FORM 1 SAMPLE Z-INTERVAL FOR µ 1 SAMPLE T-INTERVAL FOR µ 2 INDEPENDENT SAMPLE T-INTERVAL FOR µ 2 DEPENDENT SAMPLE T-INTERVAL FOR µ Z-INTERVAL FOR P SAMPLE SIZE Chapter 7 Confidence Intervals

Introduction In this chapter, we will estimate the population parameter using an interval. This interval will ‘capture’ the true population parameter with a certain measure of precision. Example: A 95% confidence interval for µ is (13, 18)

Notations and General Form CI - confidence interval CV - critical value ME - margin of error SE - standard error SD - standard deviation pt.est. - point estimate Z α/2 - normal distribution critical value(use invnorm) t (n-1,α/2) - students t distribution critical value with n-1 degrees of freedom (use math solver or the invT)

Notations and General Form General Formula:

Notations and General Form

Critical Value Z-critical value Z 1- α/2 = InvNorm(1 - α/2) example: 95% Confidence interval α = 1 – 0.95 = 0.05 Z 1- α /2 = InvNorm(1- α/2) = InvNorm(1-(.05/2)) = 1.96

Notations and General Form Critical Value T-critical value t (1-α /2, df) = InvT(1-α/2, degrees of freedom) or t (1-α /2, df) = L on math solver (for TI-83) Math -> solver -> tcdf(L,U,D) – A L = CV U = 9999 D = degrees of freedom A = α/2

Notations and General Form Example: 95% confidence interval based on a sample size of 20. (α =.05, df = n-1 = 19) T-critical value t (1-α /2, df) = InvT(1-(0.05/2), 19) Math->solver->tcdf(L,U,D) – A L = CV (highlight and then press alpha enter) U = 9999 D = 19 A =.025 ANSWER: t (1-0.025, 19) = 2.0930

1 Sample Z-interval for µ Population Standard Deviation (σ) is known

1 Sample Z-interval for µ Suppose the time allotted for commercials on a primetime TV program is known to have a normal distribution with a standard deviation of 1.5 minutes. A study of 25 showings gave an average commercial time of 11 minutes. Find the 95% confidence interval for the true population mean, μ. Given: Confidence level = 0.95 Critical value = invNorm(1-.05/2) = 1.96

1 Sample Z-interval for µ A 95% confidence interval for μ is We are 95% confident that the average time in commercials is between 10.412 and 11.588 minutes Using 1-sampZInterval in TI-83/84 Stat -> Tests -> ZInterval

Notes on intervals Effect of confidence level Higher confidence level results to a longer confidence interval since CV increases as α decreases. Example, Z = 1.645 when α =.10 while Z = 1.96 when α =.05 Effect of sample size Increasing sample size (n) shortens the confidence interval since SE = SD/sqrt(n). Note: level of significance (α) confidence level (1- α)*100%

1 Sample t-interval for µ Population Standard Deviation (σ) is unknown

1 Sample t-interval for µ A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. Assuming normal distribution for the number of words typed per minute, find a 95% confidence interval for the average number of words typed by all graduates of this school. Given: Confidence level = 0.95 Critical value = 2.201

1 Sample t-interval for µ A 95% confidence interval for μ is We are 95% confident that the average number of words the graduates type per minute is between 74.3441 and 84.2559 words. Using 1-samptInterval in TI-83/84 Stat -> Tests -> tInterval

2 Independent Sample t-interval for µ. Confidence Interval for µ 1 - µ 2 Requirement 1: Both samples where taken from a normal distribution Requirement 2: The population standard deviations are equal (σ 1 =σ 2 =σ)

2 Independent Sample t-interval for µ Do credit cards with no annual fee charge higher interest rates than cards that have annual fees? Among 29 cards surveyed, 17 had no annual fees while 12 charged an annual fee. Among the cards with no annual fee, the average interest rate was 19% (SD = 8%). Among cards with an annual fee, the average interest rate was 17% (SD = 3%). a. What assumptions do you need to get a confidence interval for the difference in average interest rate? b. Calculate the estimate of the common standard deviation. c. Construct a 95% interval estimate for the difference in average interest rates.

2 Independent Sample t-interval for µ a.What assumptions do you need to get a confidence interval for the difference in average interest rate? The samples were taken from normally distributed populations and that they have a common standard deviation.

2 Independent Sample t-interval for µ b. Calculate the estimate of the common standard deviation. No Annual FeeWith Annual Fee Average ( )0.190.170.19 – 0.17 = 0.02 SD (S)0.080.03 Sample Size (n)1712

2 Independent Sample t-interval for µ c. Construct a 95% interval estimate for the difference in average interest rates. Level of Confidence (1-α) = 0.95  α = 0.05 Critical value (t α/2,df=n1+n2-2 = t 0.05/2,df=27 = t 0.025,27 ) = 2.052 A 95% confidence interval for is calculated as Therefore, with 95% confidence, the difference in average interest rate will lie between -2.99% and 6.99%.

2 Independent Sample t-interval for µ No Annual FeeWith Annual Fee Average ( )0.190.170.19 – 0.17 = 0.02 SD (S)0.080.03 Sample Size (n)1712 Using 2-samptInterval in TI-83/84 Stat -> Tests -> 2samptInt

2 Independent Sample t-interval for µ Suppose it is of interest to estimate the difference in the first exam scores of STAT 2160 male and female students. A random sample of 16 males and 17 females taking the course this semester was drawn and asked about their scores. Among male students, the average was found to be 15.28 with a standard deviation of 2.3, while among females, the average is 16.5 with a standard deviation of 1.6. Assuming the scores follow a normal distribution with equal variances, construct a 95% CI for the true difference in the first exam scores of male and female STAT 2160 students.

2 Independent Sample t-interval for µ Step 1: Create a table of given values Step 2: Compute for the common variance MALEFEMALE n1617 Mean15.2816.5 Standard Deviation2.31.6

2 Independent Sample t-interval for µ

Step 3: Obtain the Critical Value (CV) 95% CI  α = 0.05 CV = invT(1-α/2, n 1 +n 2 -2) = invT(0.975, 31) = 2.0395

2 Independent Sample t-interval for µ Step 4: Plug-in the values in the CI formula

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2 Dependent Sample t-interval for µ There is one sample but measured twice

2 Dependent Sample t-interval for µ The table below shows the opening and closing prices of a sample of 10 active stocks on a certain day. Give an estimate of the difference in the average stock prices? What is the corresponding standard error of this estimate? Opening PriceClosing Price Intel Corp16.0916.25 Citigroup Inc17.4716.32 Bank of America Corp25.624.67 JPMorgan Chase & Co41.5242.33 General Electric Co20.1119.37 Microsoft Corp23.7723.44 Pfizer Inc17.0316.93 Exxon Mobil Corp67.6962.92 Coca-Cola Co46.4849.23 Alcoa Inc11.5410.09

2 Dependent Sample t-interval for µ First, we need to take the difference between the two prices for each stock. Opening PriceClosing PriceDifference Intel Corp16.0916.25-0.16 Citigroup Inc17.4716.321.15 Bank of America Corp25.624.670.93 JPMorgan Chase & Co41.5242.33-0.81 General Electric Co20.1119.370.74 Microsoft Corp23.7723.440.33 Pfizer Inc17.0316.930.1 Exxon Mobil Corp67.6962.924.77 Coca-Cola Co46.4849.23-2.75 Alcoa Inc11.5410.091.45

2 Dependent Sample t-interval for µ Then calculate the mean of the differences and the corresponding standard deviation then calculate the standard error.

2 Dependent Sample t-interval for µ Construct a 95% confidence interval for the difference in opening and closing stock prices. Use 1-samptInterval in TI-83/84 since we converted the two samples into a sample of differences. Answer: (-0.7875, 1.9375)

Z-interval for P

In a random sample of 500 families owning television sets in the city of Hamilton, Canada, it was found that 340 subscribed to HBO. Find a 95% confidence interval for the actual proportion of families in this city who subscribe to HBO. Given: Find: A 95% CI for the population proportion, p. Level of Confidence (1-α) = 0.95  α = 0.05 Critical value (z 1-α/2 = z 1-0.05/2 = z 0.975 ) = InvNorm(0.975) = 1.96

Z-interval for P The confidence interval is calculated as follows: Therefore, we are 95% confident that the actual proportion of families in this city who subscribe to HBO is between 64% and 72%. A function in TI-83/84 is 1propZInt. Stat -> Tests -> 1propZInt

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Sample Size In order to be (1-α) x 100% confident that the sample mean is within a distance ME of the mean μ, choose a sample size equal to n = z 2 σ 2 /M 2 For computing sample size for estimating population proportion, the formula is

Sample Size A consumer group wishes to estimate the average electric bills for the month of July for single-family homes in a large city. Based on studies conducted in other cities, the standard deviation is assumed to be \$25. The group wants to estimate the average bill for July to be within  \$5 of the true average with 95% confidence. a. How many single-family homes should be selected? b. If the group wants to be correct to within  \$10, what sample size is necessary? c. If 99% confidence and a sampling error of  \$5 are desired, how many single-family homes are necessary? Given: ME = 5 Level of Confidence (1-α) = 0.95  α = 0.05 Critical value (z α/2 = z 0.05/2 = z 0.025 ) = InvNorm(0.025) = -1.96 σ = 25

Sample Size a. How many single-family homes should be selected? Given: ME = 5 Level of Confidence (1-α) = 0.95  α = 0.05 Critical value (z α/2 = z 0.05/2 = z 0.025 ) = InvNorm(0.025) = -1.96 σ = 25

Sample Size b. If the group wants to be correct to within  \$10, what sample size is necessary?

Sample Size c. If 99% confidence and a sampling error of  \$5 are desired, how many single-family homes are necessary? M = 5 Level of Confidence (1-α) = 0.99  α = 0.01 Critical value (z α/2 = z 0.01/2 = z 0.005 ) = InvNorm(0.01/2) = -2.576 σ = 25

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Sample Size In a random sample of 500 families owning television sets in the city of Hamilton, Canada, it was found that 340 subscribed to HBO. Suppose that a CI for the proportion of families who subscribe to HBO is computed at 90% confidence level and within ± 0.05, what will the new sample size be?

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