Question

4. Multiple Choice Question We studied a characteristic of iron. Under the normality assumption of its distribution and using its sample mean and standard deviation, we computed the following confi- dence interval [2271.7688,2308.2312). This interval is a confidence interval for the mean of that characteristic at a level of confidence of 90%. We are told the the sample size is 16 and the sample mean is I = 2290. What is the value of the sample standard deviation s? A. 44.33 B. 41.6 C. 83.2 D. 54.38 E. none of the preceding I

Answer 1

Given,

Confidence interval = [2271.7688, 2308.2312]

Sample size (n) = 16

Sample mean() = 2290

$We\ know,\\\\ CI = \left (\bar{x}\pm z\times \frac{s}{\sqrt{n}} \right )$

Here, z= 1.6449~ 1.645.

$\therefore CI= \left (2290 \pm 1.645\times \frac{s}{\sqrt{16}} \right )$

$\therefore [2271.7688, 2308.2312 ] = \left (2290 \pm 1.645\times \frac{s}{\sqrt{16}} \right )$
S S :: [2271.7688, 2308.2312] = 2290 – 1.645 x 2290 + 1.645 x 16 V16
$\therefore 2271.7688= \left (2290 - 1.645\times \frac{s}{\sqrt{16}} \right )$$\therefore \left (1.645\times \frac{s}{4} \right ) = 2290 - 2271.7688$

$\therefore \left (1.645\times \frac{s}{4} \right ) = 18.2312$

$\therefore \left (s \right ) = \frac{18.2312\times 4}{1.645} = 44.33$

Hencentral, we get value of sample standard deviation (s) = 44.33