Mathematics: 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

plural form of leaf.Step-by-step explanation:...

Basic Computation: Testing M1 2 M2 A random sample

ToxicMonkey

11.02.2022

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

Step-by-step 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:

$\mathbf{H_o:\mu_1=\mu_2}$

$\mathbf{H_1:\mu_1\ne\mu_2}$

Level of significance = 0.01

Using the Z-test statistics;

$Z = \dfrac{\overline x_1 - \overline x_2}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}}$

$Z = \dfrac{10- 12}{\sqrt{\dfrac{3^2}{49} + \dfrac{4^2}{64}}}$

$Z = \dfrac{-2}{\sqrt{\dfrac{9}{49} + \dfrac{16}{64}}}$

$Z = \dfrac{-2}{\sqrt{0.18367 +0.25}}$

$Z = \dfrac{-2}{\sqrt{0.43367}}$

$Z = \dfrac{-2}{0.658536}$

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

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