Question

The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers. The data are normally distributed and s 10.048 weeks. Construct and interpret a 95% confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl. 353 31 42 36 38 27 47 38 57 25 4129 Click the icon to view the table of critical values of the chi-square distribution. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to three decimal places as needed.) O A. If repeated samples are taken, 95% of them will have the sample standard deviation between O B. There is a 95% probability that the true population standard deviation is between and OC. There is 95% confidence that the population standard deviation is between and and

Answer 1

Solution

given : s = 10.048, n = 12

Confidence level (C) = 0.95

1+0 Area to the right of XL = = 1 +0.95 -= 0.975

Area to the right of 2 - 1-C 1-0.95 -= 0.025

degrees of freedom df) = n-1 = 12 -1 = 11

Refer chi-square table or use excel formula "=CHISQ.INV.RT(0.975,11)" & "=CHISQ.INV.RT(0.025,11)" to find the critical values.

x = x3 975, 11 = 3.816

XR = xố 025, 11 = 21.920

$\mathbf{form ula:}\; confidence\; interval\; estimate\; of\; \sigma$

$\small \Rightarrow \sqrt{\frac{\left (12-1 \right )*10.048^{2}}{21.920}}<\sigma<\sqrt{\frac{\left (12-1 \right )*10.048^{2}}{3.816}}$

$\small \Rightarrow \mathbf{{\color{Red} 7.118 } <\sigma<{\color{Red} 17.060 } }$

$\small \mathbf{\mathbf{{\color{Blue} C}}.\;{\color{Red} \mathbf{ }}There\; is \;95\%\; confidence\; that \;the \;population\; standard \;deviation \;is\; between}$$\small \mathbf{{\color{Red} 7.118}\;and\;{\color{Red} 17.060}}$