If he selects a random sample of 400 workers from the factory and the sample shows an average of...

Question:

If he selects a random sample of 400 workers from the factory and the sample shows an average of $15 per hour with a standard deviation of$6, a 95% confidence interval for the average hourly pay in this factory is what?

Confidence interval

The confidence level shows the probability that the confidence interval contains the population parameter which is obtained from the sample statistics. The confidence interval is shortened as CI.

Given information

• Sample size: 400
• Sample mean: 15
• Standard deviation: 6

The critical value is obtained from the standard normal distribution at level of significance (0.05)

The Critical value is 1.96

The 95% confidence interval for the average hourly pay in this factory is calculated as follow.

{eq}\begin{align*} P\left( {\bar X - {Z_{\alpha /2}}\dfrac{\sigma }{{\sqrt n }} < \mu < \bar X + {Z_{\alpha /2}}\dfrac{\sigma }{{\sqrt n }}} \right) &= 0.95\\ P\left( {15 - 1.96\dfrac{6}{{\sqrt {400} }} < \mu < 15 + 1.96\dfrac{6}{{\sqrt {400} }}} \right) &= 0.95\\ P\left( {14.412 < \mu < 15.588} \right) &= 0.95 \end{align*}{/eq}

Interpretation: There is 95% confident that for the average hourly pay in this factory is lies between confidence limits (14.412 to 15.588)