Question

Here are the IQ test scores of 31 seventh-grade girls in a Midwest school district:

114 100 104 89 102 91 114 114 103 105 108 130 120 132 111 128 118 119 86 72 111 103 74 112 107 103 98 96 112 112 93

These 31 girls are an SRS of all seventh-grade girls in the school district. Suppose that the standard deviation of IQ scores in this population is known to be σ = 15. We expect the distribution of IQ scores to be close to Normal. Estimate the mean IQ score for all seventh-grade girls in the school district, using a 98% confidence interval.

Answer 1

Given Sample: 100, 104, 89, 102, 91, 114, 114, 103, 105, 108, 130, 120, 132, 111, 128, 118, 119, 86, 72, 111, 103, 74, 112, 107, 103, 98, 96, 112, 112, 93 We calculate sample mean Sample Mean, x = Sum M/n = 3281/31 = 105.83871 90 % CL for mu Using Normal dist Sample Mean = x = 105.83871 Population Standard deviation = sigma = 15 Sample Size = n = 31 Significance level = alpha = 1- Confidence = 1 - 0.9 = 0.1 The Critical Value = zalpha/2 = zo.os = 1.645 (From Z table using inter polation the distance between 1.64 and 1.65) Critical values plusminus Z_alpha/2 = plusminus 1.645 Margin of error + E Z_alpha/2 times sigma/squareroot n = 1.645 times 15/squareroot 31 approximately 4.431762 Margin of Error E = 4.432 Limits of 90% confidence interval are given by: Lower limit X - E = 105.83871 - 4.431762 < < 101.406948 > > 101.407 Upper limit = x + E 105.83871 + 4.431762 < < 110.270472 < < 110.270

Answer 2

Given Sample: 114, 100, 104, 89, 102, 91, 114, 114, 103, 105, 108, 130, 120, 132, 111, 128, 118, 119, 86, 72, 111, 103, 74, 112, 107, 103, 98, 96, 112, 112, 93 Σ(x) 3281 31 We calculate sample mean Sample Mea n , x = 105.83871 1 90% CI for using Normal dist Sample Mean X 105.83871 Population Standard deviation = σ = 15 Sample Sizen 31 Significance level a 1- Confidence10.9 0.1 The Critical Value Zo/2 Z0.05- 1.645 (From z table, using interpolation, 1/2 th distance between 1.64 and 1.65) Critical Values , ±h/,-±1.645 Margin of Error-E-za/2 × Margin of Error, E-4432 15 xN 4.431762 /31 Limits of 90% confidence interval are given by Lower limit X-E 105.83871-4.431762101.406948 Upper limit = x + E 105.83871 + 4.431762 110.270472 101.407 110.270 90% confidence interval is: x - (101.406948, 110.270472) E 105.83871 4.431762 90% CI using normal dist: ( 101.407) < e < (110.270 Also, the exact answer using TI-83/84 or excel is 105.83871 tinvNorm(0.95) x 15/sqrt(31) -(101.4073427407,110.2700772593) noau Thank youse