Question

A sample from 16 shortstops was obtained in an effort to determine whether shortstops have higher salaries than other players. The mean salary (in millions of dollars) for shortstops was 1.95. Assume that the population mean for all baseball players was 1.85 and the population standard deviation was 1.61. What is the 95 percent confidence interval for these data?

Answer 1

Answer:

1.16 to 2.75

Step-by-step explanation:

The mean salary (millions of dollars) for shortstops was 1.95.

$\bar{x}=1.85$

The population mean for all baseball players was 1.85

$\mu = 1.85$

The population standard deviation was 1.61.

$\sigma = 1.61$

Formula of confidence interval :$x-z\times \frac{\sigma}{\sqrt{n}}$ to $x+z\times \frac{\sigma}{\sqrt{n}}$

z at 95% confidence level = 1.96

Substitute the values in the formula

Confidence interval :$1.95-1.96\cdot\frac{1.61}{\sqrt{16}}$ to $1.95+1.96\cdot\frac{1.61}{\sqrt{16}}$

Confidence interval : 1.16 to 2.75

Hence the 95 percent confidence interval for these data is 1.16 to 2.75