Question

Refer to the Scheer Industries example in Section 8.2. Use 6.84 days as a planning value for the population standard deviation. a. Assuming 95% confidence, what sample size would be required to obtain a margin of error of 1.5 days (to the nearest whole number)? b. Assuming 90% confidence, what sample size would be required to obtain a margin of error of 2 days (to the nearest whole number)?

Answer 1

Answer: a) 80  b) 32

Step-by-step explanation:

a) Given : Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Standard deviation : s =6.84

Margin of error : E= 1.5

The formula to find the sample size is given by :-

$n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2$

i.e. $n=(\dfrac{(1.96)(6.84)}{1.5})^2=79.88069376\approx80$

Hence, the required minimum sample size = 80

b) Given : Significance level : $\alpha=1-0.90=0.10$

Critical value : $z_{\alpha/2}=1.645$

Standard deviation : s =6.84

Margin of error : E= 2

The formula to find the sample size is given by :-

$n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2$

i.e. $n=(\dfrac{(1.645)(6.84)}{2})^2=31.65075081\32$

Hence, the required minimum sample size =32