Question

A random sample of n1 = 49 measurements from a population with population standard deviation σ1 = 3 had a sample mean of x1 = 9. An independent random sample of n2 = 64 measurements from a second population with population standard deviation σ2 = 4 had a sample mean of x2 = 11. Test the claim that the population means are different. Use level of significance 0.01.

Compute the corresponding sample distribution value. (Test the difference μ1 − μ2. Round your answer to two decimal places.)

Answer 1

The answer is "-3.04"

Explanation:

`\to \bar{x_1}-\bar{x_2}=9-11=-2`

Sample distribution:

`z=\frac{\bar{x_1}-\bar{x_2}- \bar{\mu_1}-\bar{\mu_2}}{\sqrt{(\sigma_(1)^2)/(n_1)+(\sigma_(2)^2)/(n_2)}}\\\\`

`=\frac{(-2)-0}{\sqrt{(3^2)/(49)+(4^2)/(64)}}\\\\=\frac{-2}{\sqrt{(9)/(49)+(16)/(64)}}\\\\=\frac{-2}{\sqrt{(576+784)/(3136)}}\\\\=\frac{-2}{\sqrt{(1360)/(3136)}}\\\\=(-2)/(√(0.433))\\\\=(-2)/(0.658)\\\\=-3.039\\\\=-3.04`