Answer:
There is sufficient evidence to support the claim.
Step-by-step explanation:
Given
[tex]\alpha = 0.10[/tex]
Make A
[tex]\bar x_1 = 41[/tex]
[tex]\sigma_1= 4.6[/tex]
[tex]n = 35[/tex]
Make B
[tex]\bar x_2 = 45[/tex]
[tex]\sigma_2 = 4.3[/tex]
[tex]n = 35[/tex]
First, we state the null and alternate hypothesis
[tex]H_0 : \mu_1 = \mu_2[/tex]
[tex]H_1 : \mu_1 \ne \mu_2[/tex]
Next, calculate test statistic:
[tex]z = \frac{(\bar x_1 - \bar x_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}[/tex]
[tex]z = \frac{(41- 45)-0}{\sqrt{\frac{4.6^2}{35}+\frac{4.3^2}{35}}}[/tex]
[tex]z = \frac{-4}{\sqrt{0.6046+0.5283}}[/tex]
[tex]z = \frac{-4}{\sqrt{1.1329}}[/tex]
[tex]z = \frac{-4}{1.0643}[/tex]
[tex]z = -3.7583[/tex]
The critical value at [tex]\alpha = 0.10[/tex] is: [tex]\±1.282[/tex]
[tex]-3.7583 < -1.282[/tex]
So, we reject [tex]H_0[/tex]
This implies that; there is sufficient evidence to support the claim.
