Answer:
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 [tex]n_1[/tex] = 49
Population standard deviation [tex]\sigma_1[/tex] = 3
Sample mean [tex]\overline x _1[/tex]= 10
Second population:
Sample size [tex]n_2[/tex] = 64
Population standard deviation [tex]\sigma_2[/tex]= 4
Sample mean [tex]\overline x_2[/tex]= 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:
[tex]\mathbf{H_o:\mu_1=\mu_2}[/tex]
[tex]\mathbf{H_1:\mu_1\ne\mu_2}[/tex]
Level of significance = 0.01
Using the Z-test statistics;
[tex]Z = \dfrac{\overline x_1 - \overline x_2}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}}[/tex]
[tex]Z = \dfrac{10- 12}{\sqrt{\dfrac{3^2}{49} + \dfrac{4^2}{64}}}[/tex]
[tex]Z = \dfrac{-2}{\sqrt{\dfrac{9}{49} + \dfrac{16}{64}}}[/tex]
[tex]Z = \dfrac{-2}{\sqrt{0.18367 +0.25}}[/tex]
[tex]Z = \dfrac{-2}{\sqrt{0.43367}}[/tex]
[tex]Z = \dfrac{-2}{0.658536}[/tex]
Z = - 3.037
Z [tex]\simeq[/tex] - 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 [tex]H_o[/tex]
