Answer to Question #349289 in Statistics and Probability for jai

Question

a random sample of 81 items is taken, producing a sample mean of 47. The population standard deviation is 5.89. construct a 90 confidence interval to estimate the population mean.

Answer 1

The critical value for $\alpha = 0.1$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The corresponding confidence interval is computed as shown below:

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

$$=(47-1.6449\times\dfrac{5.89}{\sqrt{81}},47+1.6449\times\dfrac{5.89}{\sqrt{81}})$$

$$=(45.9235, 48.0765)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $45.9235 < \mu < 48.0765 ,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(45.9235, 48.0765).$