Question

1 2 3 5 6 The scores on an examination in psychology are approximately normally distributed with mean 500 and an unknown standard deviation. The following is a random sample of scores from this examination 410,438, 469, 501, 527, 539,586 Find a 90% confidence interval for the population standard deviation. Then complete the table below Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval? X PE

Answer 1

ANSWER:

+(501-495.71)2+ (527-495-71)+(539-495.715 Given, random sample of scores - 410,438,469,501,527,539,586 &=0.05 Sample n = 7 410 + 438+469+501 +597 +539+586 mean 7 M = 495.71 Standard deviation E(X; -M) 2 n ( 410 – 495.71):7 (438–495.71) + (469-495:-713) +(586-495.71)] 7 7346.20 +3330.44 + 713.42 +68.72+ 979.06+1874.02 +8152.28 7 22,464-14 7 s = 56.64

x² (alpha/2, n-1) = x²(0.05/2,7-1) - X?(0.025,6) 14.449 (:chi-square table) X? (1-alphal 2, n-1) = x2(1-0.0572 , 7-1) =x² (0.975,6) 1.237 : confidence interval of 90% = (n-1) 52 (n-1)5² x?calphal2 , n-1) coe X? (1-alphal2.n-1) (1-1) 156.64² (7-1)(56-6402 Lok 14.449 1-237 6X3208.08 6x3208:08 co z 14.449 1-237 19248.53 14.449 19248.53 1.237 26 Z. 11,332.17092 15,560.65 36.49 co2 124.74

3. .: Lower limit 36.49 upper limit 124.74