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Question 1 of 3, Step 4 of 5 1/15 Correct The following data give the number of hours 5 students spent studying and their corresponding grades on their midterm exams. Hours Studying 03 4 4 4 Midterm Grades 76 78 80 90 92 Table Copy Data Step 4 of 5 : Construct the 95% confidence interval for the slope. Round your answers to three decimal places. Answer How to enter your answer Tables Keypad Previous Answers Lower endpoint Upper endpoint

Answer 1

X Y XY XX YY 0 76 0 0 5776 3 78 234 96084 4 80 320 16 6400 4 90 360 16 8100 4 92 368 16 8464 X = 15.0, y = 416.0 XY = 1282.0, XX = 57.0, YY = 34824.0


$\\SST = \sum Y^2 - \frac{(\sum Y)^2}{n} \\SST = 34824.0 - \frac{(416.0)^2}{5} \\SST = 34824.0 - \frac{173056.0}{5} \\SST = 34824.0 - 34611.2 \\SST = 212.8 \\SSR = \frac{(\sum XY - \frac{\sum X\sum Y}{n})^2}{\sum X^2 - \frac{(\sum X)^2}{n}} \\SSR = \frac{(1282.0 - \frac{15.0(416.0)}{5})^2}{57.0 - \frac{(15.0)^2}{5}} \\SSR = \frac{(1282.0 - 1248.0)^2}{57.0 - 45.0} \\SSR = \frac{1156.0}{12.0} \\SSR = 96.333 \\SSE = SST - SSR \\SSE = 212.8 - 96.333 \\SSE = 116.47$

$\\\hat{\sigma}^2 = \frac{SS_E}{n-2} \\\hat{\sigma}^2 = \frac{SS_T - SS_R}{n-2} \\\hat{\sigma}^2 = \frac{212.8-96.33}{5-2} \\\hat{\sigma}^2 = \frac{116.47}{3} \\\hat{\sigma}^2 = 38.822$

Confidence Interval (in %) = 95

Ξ Sus = 0.05and d. f. = 1 – 2 = 3 ta/2,n-2 = to025,3 = 3.1824 β, ηΣΧΥ – ΣΧ *ΣΥ ηΣΧ2 – (ΣΧ)2 5 * 1282.0 – 15.0 * 416.0 β & 2.8333 5 * 57.0 – 15.0 * 15.0 σ2 S.E.(B1) = standard error of ĝi Where, σ? = σ? = 38.82 (ΣΧ)2 And, Sir = ΣΧ? - (15.0)2 Sir = 57.0 – 5 Sx = 12.0 38.822 So, S.E.(β) = 12.0 S.Ε. (β) = 1.7987 n

Since we know that

B1 – ta/2,1–2 * S.E.(B1) < B1 < B1 + ta/2,1–2 * S.E.(B1) 2.8333 – 3.1824 * 1.7987 < B1 < 2.8333 +3.1824 * 1.7987 2.8333 – 5.7242 < Bu < 2.8333 +5.7242 – 2.8909 < B1 < 8.5575

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