Question

Enough money has been budgeted to collect independent random samples of size $n_1=n_2=100$from populations 1 and 2 to estimate localid="1664867109106" ${\mu }_1-{\mu }_2$. Prior information indicates that ${\sigma }_1={\sigma }_2=10$. Have sufficient funds been allocated to construct a 90% confidence interval for${\mu }_1-{\mu }_2$of width 5 or less? Justify your answer.

Short answer

The required sample size is 22. Since, the sample size for each population is large enough.

Step-by-step solution

Given Information

The sample size of two populations are given below

$n_1=n_2=100$

The standard deviation of two populations are given below

${\sigma }_1={\sigma }_2=10$

The sampling error is

$SE=5$

Z-value

For $\alpha =0.1$

The z-value is given by

$\begin{array}{rcl}{z}_{\frac{\alpha }{2}}& =& z_{0.05}\\ & =& 1.645\end{array}$

Compute the sample

For, $z=1.645,{\sigma }_1={\sigma }_2=10 andSE=5$

The sample is calculated as

$\begin{array}{rcl}{n}_1& =& n_2\\ & =& \frac{{\left(z_{0.05}\right)}^2\times \left[{\sigma }_1^2+{\sigma }_2^2\right]}{{\left(SE\right)}^2}\\ & =& \frac{{1.645}^2\times \left[{10}^2+{10}^2\right]}{5^2}\\ & =& \frac{2.7060\times 200}{25}\\ & =& \frac{541.2}{25}\\ & =& 21.648\\ & \approx & 22\end{array}$

Therefore, the required sample size is 22.

Since, the sample size for each population is large enough.