Question

A random sample of size 36 was taken from a population distributed as Nμ,3.92.The value of the sample x was 15.6. i. Find a 90% confidence interval for μ. (5mks)

It is believed that value of μ is 17. Use your confidence interval to comment on this belief.(2mks)

Answer 1

a)

CI = x ± z*$\frac{\sigma}{\sqrt{n}}$

Given that x=15.6, $\sigma$=3.92, n=36

From the normal distribution table, the critical value for 90% confidence interval z= ±1.645

Then;

90% CI = 15.6 ±1.96*$\frac{3.92}{\sqrt{36}}$

The lower limit = 15.6 - 1.28 =14.32

The upper limit = 15.6 + 1.28 = 16.88

Therefore, the 90% confidence interval for population mean is (14.32, 16.88)

b) Since the interval does not contain the value of $\mu$ =17, we can conclude that the value of the population mean $\mu$ $\neq$ 17