Answer: between 56.45% and 57.55%
Explanation:
We are given:
μ = population mean = 57%
σ = population standard deviation = 3.50%
n = sample size = 40
CL = confidence level = 68% = 0.68
We need to calculate the confidence interval in which we should find the sample mean \(\bar{x}\).
First of all, we need to calculate the sample standard deviation:
\(s = \frac{\sigma}{\sqrt{n} } = \frac{3.50}{\sqrt{40} }\)
s = 0.55
Now, there are two equivalent ways to arrive to the answer:
1) short way: use the Empirical Rule, also known as 68–95–99.7 Rule, to remember that 68% of the data lies within 1 standard deviation from the mean.
Therefore, the confidence interval should be:
\(\mu - 1 s \leq \bar{x} \leq \mu + 1 s\)
57 - 0.55 ≤ \(\bar{x}\) ≤ 57 + 0.55
56.45 ≤ \(\bar{x}\) ≤ 57.55
2) long way: make all the calculations.
First, find the critical value corresponding to the confidence level required:
\(1 - \frac{\alpha }{2} = 1 - \frac{1 - CL}{2} = 1 - \frac{1 - 0.68}{2} = 0.84\)
Now, since the sample size is greater than 30 and it is said to be statistical significant, we can use a z-score table (instead of a t-score table).
Looking for which z-score corresponds to a probability closest to 0.84, we get z = 1.00 (which confirms the Empirical Rule).
Now, we can find the confidence interval requested:
\(\mu - z \cdot s \leq \bar{x} \leq \mu + z \cdot s\)
57 - 1·0.55 ≤ \(\bar{x}\) ≤ 57 + 1·0.55
56.45 ≤ \(\bar{x}\) ≤ 57.55
