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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.)

Respuesta :

Answer:

The answer is "-3.04"

Step-by-step explanation:

[tex]\to \bar{x_1}-\bar{x_2}=9-11=-2[/tex]

Sample distribution:

[tex]z=\frac{\bar{x_1}-\bar{x_2}- \bar{\mu_1}-\bar{\mu_2}}{\sqrt{\frac{\sigma_{1}^2}{n_1}+\frac{\sigma_{2}^2}{n_2}}}\\\\[/tex]

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

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