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Basic Computation: Testing M1 2 M2 A random sample of 49 measurements from a population with population standard deviation 3 had a sample mean of 10. An independent random sample of 64 measurements from a second population with population standard deviation 4 had a sample mean of 12. Test the claim that the population means are different. Use level of significance 0.01. (a) Check Requirements What distribution does the sample test statistic follow

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Answer:

The sample statistics follows a standard normal distribution since the sample size are large enough.

Explanation:

Given that:

First population:

Sample size
n_1 = 49

Population standard deviation
\sigma_1 = 3

Sample mean
\overline x _1= 10

Second population:

Sample size
n_2 = 64

Population standard deviation
\sigma_2= 4

Sample mean
\overline x_2= 12

The sample statistics follows a standard normal distribution since the sample size are large enough.

The null and alternative hypotheses can be computed as:


\mathbf{H_o:\mu_1=\mu_2}


\mathbf{H_1:\mu_1\\e\mu_2}

Level of significance = 0.01

Using the Z-test statistics;


Z = \frac{\overline x_1 - \overline x_2}{\sqrt{(\sigma_1^2)/(n_1) + (\sigma_2^2)/(n_2)}}


Z = \frac{10- 12}{\sqrt{(3^2)/(49) + (4^2)/(64)}}


Z = \frac{-2}{\sqrt{(9)/(49) + (16)/(64)}}


Z = (-2)/(√(0.18367 +0.25))


Z = (-2)/(√(0.43367))


Z = (-2)/(0.658536)

Z = - 3.037

Z
\simeq - 3.04

The P-value = 2P (z < -3.04)

From the z tables

= 2 × (0.00118)

= 0.00236

Thus, since P-value is less than the level of significance, we fail to reject the null hypotheses
H_o

User Margusholland
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