Question

A sample of size n = 100 produced the sample mean of ܺത= 16. Assuming the population standard deviation σ= 3, compute a 95% confidence interval for the population mean μ. (b) Assuming the population standard deviation σ= 3, how large should a sample be to estimate the population mean μ with a margin of error not exceeding 0.5 with a 95% confidence interval?

Answer 1

Answer:

a. Confidence Interval =[15.412,16.588]

b $n\geq 139$

Step-by-step explanation:

a. Given a sample  of size n=100 and mean =16, standard deviation =3, the 95% confidence interval is calculated using the formula;

$\bar X\pm z\frac{\sigma}{\sqrt{n}}$

#We substitute for the Lower bound interval;

$=\bar X- z\frac{\sigma}{\sqrt{n}}\\\\\\=16-1.96\times\frac{3}{\sqrt{100}}\\\\\\=15.412$

#For the upper bound;

$=\bar X+ z\frac{\sigma}{\sqrt{n}}\\\\=16+1.96\times \frac{3}{\sqrt{100}}\\\\\\=16.588$

Hence, the 95% confidence interval for the population mean is between 15.412 and 16.588

b. To find how large the sample has to be, we calculate using the formula;

$n\geq (z_{\alpha/2 }\sigma/\bigtriangleup)^2$

where;

• $\bigtriangleup$ is the required margin, 0.5

• $\alpha=0.05$ and $z_{\alpha/2}=1.96$

#We substitute in the equation;

$n\geq (\frac{1.96\times 3}{0.5})^2\\\\=138.298\approx 139$

Hence, the desired sample size is atleast 139