Question

A random sample of 40 regions give a sample mean of × = R688 per 50 kilograms of watermelon. Assume that the population standard deviation is R192 per so kilograms. The critical value to estimate a 95% confidence interva of the population mean
1. 1, 645
2. 1.684
3. 2. 021
4. 1.96
5. 5%

Answer 1

Since the population standard deviation is known and the sample is greater than 30, Normal distribution is used.

$Cv=Z_{\frac{\alpha}{2}}$

$Z_{0.025}=1.96$

Thus, Option 4 is correct.