Answer to Question #222444 in Statistics and Probability for bcd

Question

A food manufacturer wishes to estimate the average time required to process orders

received from supermarkets. A random sample of 60 orders revealed a mean time of 2.4

days and a standard deviation of 1.2 days.

i) Compute a 95% confidence interval for the population mean process time.

ii) If the manufacturer wishes to be within 0.2 days of the true average time, what size

sample should be drawn?

Answer 1

i) The critical value for $\alpha=0.05$ and $df=n-1=60-1=59$ degrees of freedom is $t_c= 2.000995.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(2.4-2.000995\times\dfrac{1.2}{\sqrt{60}}, 2.4+2.000995\times\dfrac{1.2}{\sqrt{60}})$$

$$=(2.09, 2.71)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $2.09<\mu< 2.71,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(2.09, 2.71).$

ii)

$$t_c\times\dfrac{s}{\sqrt{n}}\leq0.2$$

$$n\geq(\dfrac{t_c\times s}{0.2})^2$$

$$n\geq(\dfrac{t_c\times 1.2}{0.2})^2$$

$$n\geq 36(t_c)^2$$

$n=141, df=140, t_c=1.977054$

$$36(1.977054)^2=140.7$$

$n=140, df=139, t_c=1.977178$

$$36(1.977178)^2=140.7$$

Hence

$$n=141$$