Question

A survey of 98 randomly selected students finds that they spend a mean of $435 per semester. Assume that the population standard deviation is $56 per month.

Confidence Interval: What is the 95% confidence interval to estimate the population mean? (Round your answers to two decimal places.)

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Answer 1

The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\dfrac{\sigma}{\sqrt{n}})$$

$$=(435-1.96\dfrac{56}{\sqrt{98}},435+1.96\dfrac{56}{\sqrt{98}})$$

$$=(423.913, 446.087)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $423.913 < \mu < 446.087,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(423.913, 446.087).$