Download Statistical Analysis of Professor Salaries: Confidence Intervals and Hypothesis Testing - and more Study notes Statistics in PDF only on Docsity! Fitzpatrick 2 1. The sample mean is from a random sample, the population distribution is normal and the population standard deviation is unknown, therefore to construct a 95% Confidence interval for µf and µm I will use the following formula: ± (t critical value) (s / √n) a) For µf , where n = 14, f = 21357.14, sf = 6151.873, dff = 13, t critical value = 2.160 A 95% Confidence interval for µf is (17805.760, 24908.520). The range of the confidence interval is defined by the sample statistic + margin of error. And the uncertainty is denoted by the confidence level. Therefore, this 95% confidence interval says that the female professor mean salary falls within the interval $17,805.760 and $24,908.520. That is we are 95% confident that the true mean lies between this range. b) For µm where n = 38, m = 24696.79, sm =5646.409, dfm = 37, t critical value = 2.026 A 95% Confidence interval for µm is (22841.038, 26552.542). The range of the confidence interval is defined by the sample statistic + margin of error. And the uncertainty is denoted by the confidence level. Therefore, this 95% confidence interval says that the male professor mean salary falls within the interval $22,841.038 and $26,552.542. That is that we are 95% confident that the true mean lies between this range. 2. 1. Population characteristic of interest µ =mean salary for all professors 2. Null hypothesis H0: µ = 26,000 3. Alternate Hypothesis Ha: µ <26,000 4. Significance level: α = .05 5. Test statistic: t= x́−hypothesized value s √n 6. Assumptions: The population is normal and the standard deviation is unknown, therefore the use of the t test is reasonable. 7. Computations: Where n = 52, = 23797.654, s = 42257.897, df = 52-1 = 51, SE = s / √n = 5860.116, t =(23797.654-26000)/5860.116 =- 0.376. Fitzpatrick 5 5. 0 5 10 15 20 25 30 0 5000 10000 15000 20000 25000 30000 35000 40000 f(x) = 752.8 x + 18166.15 Salary by Years Employed Salary Linear (Salary ) Year Sa la ry Let y denote the dependent variable, Salary and x denote the independent variable, years. a) The summary statistics of the data are: n = 52 ∑ i=1 n x i = 389 ∑ i=1 n yi = 1237478 ∑ i=1 n ❑x i 2 =4457 ∑ i=1 n x i y i= 10421851 ∑ i=1 n yi 2= 31234802944 Fitzpatrick 6 ∑ i=1 n x i ∑ i=1 n yi= 481378942 Then we have: Sxy = ∑ i=1 n x i y i− (∑ i=1 n x i)(∑ i=1 n y i) n = 1164564 Sxx = ∑ i=1 n x i 2 − (∑ i=1 n xi) 2 n = 1546.981 = 389/52 = 7.481 ý = 1237478/52 = 23797.654 β̂=¿ Sxy Sxx = 1164564 1546.981 =752.798 α̂= ý− β̂ = 23797.654 – (752.798)(7.481) = 18165.972 The equation of the estimated regression line is then: ŷ= α̂+ β̂=18165.972+752.798 x b) y is expected to increase by β̂ units for each1 unit increase in x, here Salary is to increase $752.8 for each increase in years worked. c) If the year was to be 18, we would have: ŷ= α̂+ β̂=18165.972+752.798(18) = $31,719.61
