Given the population standard deviation is known as R500. A random sample of 50 individuals resulted in a mean of R10000. What is the width of the 90% confidence interval?

Answer 1

To calculate the width of the 90% confidence interval for the population mean, we'll use the formula for the confidence interval when the population standard deviation is known. The formula is:

$$CI=\overline{x}\pm z\left(\frac{\sigma }{\sqrt{n}}\right)$$

Where:

• $\overline{x}$ = sample mean
• $z$ = z-score corresponding to the desired confidence level
• $\sigma$ = population standard deviation
• $n$ = sample size

1. Determine the parameters:

   • Sample mean, $\overline{x}=R10000$
   • Population standard deviation, $\sigma =R500$
   • Sample size, $n=50$

2. Find the z-score for a 90% confidence level:

   • For a 90% confidence interval, the z-score (which corresponds to the critical value) can be found using standard normal distribution tables. The z-score for 90% confidence level is approximately $1.645$.

3. Calculate the standard error (SE): $$SE=\frac{\sigma }{\sqrt{n}}=\frac{500}{\sqrt{50}}\approx \frac{500}{7.0711}\approx 70.71$$

4. Calculate the margin of error (ME): $$ME=z\cdot SE=1.645\cdot 70.71\approx 116.67$$

5. Calculate the width of the confidence interval: The width of the confidence interval is $2\times ME$: $$\text{Width}=2\times 116.67\approx 233.34$$

Thus, the width of the 90% confidence interval is approximately R233.34.