Question

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. College students' annual earnings: 98% confidence; n = 9, sample mean is $4121, s = $873

Group of answer choices:

a) $551 < σ < $1925

b) $590 < σ < $1672

c) $531 < σ < $1708

d) $687 < σ < $1139

Answer 1

Solution Let us Consider 0 given information: Mean (x ) -$4121 Size (1) ► de- 0428 (5) Standard deviation : $ 873 Significan level [Given 99%. C8 =1-0.98 = 0.02 Since (n +)52 follicws chi-squared coith (n=) degree of freedom (n-128" => pxelles X2/2 xxx)=(1- + 781 98% CI is : (0-1S <o< (n-15? X /2 ,df x^2/2, of (914873) x 0.02./2, < 8 xi-002,8 8(762129) (8)(762129) X1 -0.01,8 x 6.01,8 6097032 6097032 **0.01,8 X0.99,8

By using chirsquare table 2 x20.01,8 20.090 x2 0.99,8 = 1.646 6097032 6097032 20.090 s 1.646 303485.913 < =) < 3764150.67 < 1924.6 550.8 551 sa 1925 option (A) is correct 55) Zo < 1925 x