* Point estimate for P is: x n
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1 Estimation and Confidence Interval Estimation and Confidence Interval: Single Mean: To find the confidence intervals for a single mean: 1- X ± ( Z 1 σ n σ known S - X ± (t 1,n 1 n σ unknown Estimation and Confidence Interval: Two Means To find the confidence intervals for two means: 1- (X 1 X ± (Z 1 σ 1 + σ n 1 n σ 1 and σ known - (X 1 X ± (t 1,n 1+n Sp 1 n n σ 1 and σ unknown Sp = S 1 (n S (n 1 n 1 + n Estimation and Confidence Interval: Single Proportion * Point estimate for P is: x n * Interval estimate for P is: p ± (Z 1 p q n Estimation and Confidence Interval: Two Proportions * Point estimate for P 1 P = p 1 p = x 1 n 1 x n * Interval estimate for P 1 P is: (p 1 p ± (Z 1 p 1q 1 + p q n 1 n 60
2 a. Find the upper limit of 95% of the confident interval for μ σ =, X = 4.5, n = 49 95% = 0.05 Z 1 = Z = 1.96 X + ( Z 1 σ n = ( = 5.06 b. Find the lower limit of 95% of the confident interval for μ X (Z 1 σ n = 4.5 ( = 3.94 σ = 5, X = 390, n = 49 a. find Z : Z = 1.96 b. find a point estimate for μ μ = X = 390 c. Find the upper limit of 95% of the confident interval for μ X + (Z 1 σ 5 = (1.96 n 49 = d. Find the lower limit of 95% of the confident interval for μ X (Z 1 σ 5 = 390 (1.96 n 49 =
3 A sample of 16 college students were asked about time they spent doing their homework. It was found that the average to be 4.5 hours. Assuming normal population with standard deviation 0.5 hours. (1 The point estimate for is: (A 0 hours (B 10 hours (C0.5 hours (D 4.5 hours ( The standard error of is: (A 0.15 hours (B 0.66 hours (C 0.06 hours (D 0.45hours (3 The correct formula for calculating confidence interval for µ is: (A (C (B (D (4 The upper limit of 95% confidence interval for µ is: (A (B (C 4.83 (D (5 The lower limit of 95% confidence interval for µ is: (A (B (C 3.63 (D 4.55 (6 The length of the 95% confidence interval for µ is: (A 4.74 (B 0.49 (C 0.83 (D
4 Estimation and Confidence Interval: Two Means To find the confidence intervals for two means: 1- (X 1 X ± (Z 1 σ 1 + σ n 1 n - (X 1 X ± (t 1,n 1+n Sp 1 n n Sp = S 1 (n S (n 1 n 1 + n Theard 1 n 1 = 0, X 1 = 7.8, σ 1 = 6.8 Thread n = 5, X = 64.4, σ = % = 0.0 Z 1 = Z 0.99 =.33 (X 1 X ± (Z 1 σ 1 + σ n 1 n ( ± ( (1: The lower limit = 3.65 (: The upper limit = ( 3.65,
5 First sample Second sample Sample size (n 1 14 Sample mean (X Sample variance (S 4 5 (1 Estimate the difference μ 1 μ 1 : E(X 1 X = X 1 X = = 0.5 ( Find the pooled estimator Sp : S p = S 1 (n 1 1+S (n 1 n 1 +n = 4(11+5(13 4 = 4.54 Sp =.13 (3 The upper limit of 95% confidence interval for µ is: 95% = 0.05 t 1,n 1+n = t 0.975,4 =.064, (X 1 X + (t 1,n 1+n Sp 1 n n (0.5 + ( =.3 (4 The lower limit of 95% confidence interval for µ is: (X 1 X (t 1,n 1+n Sp 1 n n (0.5 ( =
6 A researcher was interested in comparing the mean score of female students, with the mean score of male students in a certain test. Assume the populations of score are normal with equal variances. Two independent samples gave the following results: Female Male Sample size Mean Variance (1 The point estimate of is: (A.63 (B -.37 (C.59 (D 0.59 ( The estimate of the pooled variance ( is: (A (B (C (D (3 The upper limit of the 95% confidence interval for is : (A (B 7.55 (C (D 8. (4 The lower limit of the 95% confidence interval for is : (A (B (C (D
7 Estimation and Confidence Interval: Single Proportion * Point estimate for P is: x n * Interval estimate for P is: p ± (Z 1 p q n Solution (1: n = 00 & x = 15 (: p = x n = 15 = q = % = 0.05 Z 1 = Z = 1.96 p ± (Z 1 p q n The 95% confidence interval is: (0.038, 0.11 = ± ( A researchers group has perfected a new treatment of a disease which they claim is very efficient. As evidence, they say that they have used the new treatment on 50 patients with the disease and cured 5 of them. To calculate a 95% confidence interval for the proportion of the cured. 1. The point estimate of p is equal to: (A 0.5 (B0.5 (C 0.01 (D The reliability coefficient (Z 1 is equal is : (A 1.96 (B (C.0 (D The 95% confidence interval is equal to: (A (0.1114, (B (0.3837, (C (0.1614, (D (0.3614,
8 Estimation and Confidence Interval: Two Proportions * Point estimate for P 1 P = p 1 p = x 1 n 1 x n * Interval estimate for P 1 P is: (p 1 p ± (Z 1 p 1q 1 + p q n 1 n Solution (1 n 1 = 100 x 1 = 15 p 1 = = 0.15 q 1 = = 0.85 n = 00 x = 0 p = 0 00 = 0.10 q 1 = = 0.90 p 1 p = = 0.05 ( 95% = 0.05 Z 1 = Z = 1.96 (p 1 p ± (Z 1 p 1q 1 + p q n 1 n = (0.05 ± (1.96 (0.15( (0.1( = 0.05 ± ( The 95% confidence interval is: ( 0.031,
9 In a first sample of 100 store customers, 43 used a MasterCard. In a second sample of 100 store customers, 58 used a Visa card. To find the 95% confidence interval for difference in the proportion (p 1 p of people who use each type of credit card? 1. The value of is : (A 0.95 (B 0.5 (C 0.05 (D The upper limit of 95% confidence interval for the proportion difference is: n 1 = 100 x 1 = 43 p 1 = = 0.43 q 1 = = 0.57 n = 100 x = 58 p = = 0.58 q = = 0.4 (p 1 p + (Z 1 p 1q 1 + p q n 1 n = ( (1.96 (0.43( (0.58( (A (B (C (D The lower limit of 95% confidence interval for the proportion difference is: (p 1 p (Z 1 p 1q 1 + p q n 1 n = (0.05 (1.96 (0.15( (0.1( (A 0.78 (B (C 0.41 (D
10 Biostatistics - STAT 145 Department of Statistics Summer Semester 1431/143 Critical Values of the t-distribution (t ν=df t 0.90 t 0.95 t t 0.99 t King Saud University 91 Dr. Abdullah Al-Shiha
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