Suppose a psychologist specializing in learningdisorders wanted to estimate the mean IQ score forchildren with a particular type of learning disorder. Sheobtained a random sample of children with thislearning disorder and recorded the following IQ scores.105, 96, 102, 95, 91, 98, 109, 122, 88, 129The psychologist was willing to assume that thedistribution of IQ scores for all children with thisparticular type of learning disorder is approximatelynormal with a standard deviation equal to the standarddeviation in the general population, 11.2 points.1. Calculate the confidence interval (lower,upper) with a 90%, 95%, 99% confidencelevel (3 different intervals).2. Suppose that two other children who bothscored an IQ of 100 is added into thesample. Calculate the confidence intervalwith a 90%, 95%, 99% confidence level.3. Suppose that another set of 10 childrenwas added to the original set (so a total of20 in the sample) and their scores were:131, 124, 111, 107, 104, 100, 98, 97, 93, 90.Calculate the confidence interval with a90%, 95%, 99% confidence level.4. If we want to make the interval smaller,there are two variables we consider. Whatare the two?

Note: Since, we only answer up to 3 sub-parts, we'll answer the first 3. Please resubmit the question and specify the other sub-parts (up to 3) you'd like answered. Given: Population standard deviation σ = 11.2 Formula Used: Confidence Interval = X ± Zα/2×σnSolution: 1. Sample size n = 10 Sample mean X = 105+96+102+.....................+12910 = 103510 Sample mean X = 103.5 For 90% Confidence interval: Confidence level = 90% Significance level = 0.10 Z0.10/2 = 1.645 90% Confidence interval = X ± Zα/2×σn = 103.5 ± 1.645×11.210 = 103.5 ± 5.8262 = (97.6738, 109.3262) For 95% Confidence interval: Confidence level = 95% Significance level = 0.05 Z0.05/2 = 1.96 95% Confidence interval = X ± Zα/2×σn = 103.5 ± 1.96×11.210 = 103.5 ± 6.9418 = (96.5582, 110.4418) For 99% Confidence interval: Confidence level = 99% Significance level = 0.01 Z0.01/2 = 2.576 99% Confidence interval = X ± Zα/2×σn = 103.5 ± 2.576×11.210 = 103.5 ± 9.1236 = (94.3764, 112.6236)2. Sample size n = 12 Sample mean X = 100+100+105+96+102+.....................+12912 = 123512 Sample mean X = 102.9167 For 90% Confidence interval: Confidence level = 90% Significance level = 0.10 Z0.10/2 = 1.645 90% Confidence interval = X ± Zα/2×σn = 102.9167 ± 1.645×11.212 = 102.9167 ± 5.3186 = (97.5981, 108.2353) For 95% Confidence interval: Confidence level = 95% Significance level = 0.05 Z0.05/2 = 1.96 95% Confidence interval = X ± Zα/2×σn = 102.9167 ± 1.96×11.212 = 102.9167 ± 6.3370 = (96.5797, 109.2537) For 99% Confidence interval: Confidence level = 99% Significance level = 0.01 Z0.01/2 = 2.576 99% Confidence interval = X ± Zα/2×σn = 102.9167 ± 2.576×11.212 = 102.9167 ± 8.3286 = (94.5881, 111.2453)3. Sample size n = 20 Sample mean X = 131+124+111.....................+12920 = 209020 Sample mean X = 104.5 For 90% Confidence interval: Confidence level = 90% Significance level = 0.10 Z0.10/2 = 1.645 90% Confidence interval = X ± Zα/2×σn = 104.5 ± 1.645×11.220 = 104.5 ± 4.1197 = (100.3803, 108.6197) For 95% Confidence interval: Confidence level = 95% Significance level = 0.05 Z0.05/2 = 1.96 95% Confidence interval = X ± Zα/2×σn = 104.5 ± 1.96×11.220 = 104.5 ± 4.9086 = (99,5914 109.4086) For 99% Confidence interval: Confidence level = 99% Significance level = 0.01 Z0.01/2 = 2.576 99% Confidence interval = X ± Zα/2×σn = 104.5 ± 2.576×11.220 = 104.5 ± 6.4513 = (98.0487, 110.9513)