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To compare the dry braking distances from 30 to 0 miles per hour for two makes of​ automobiles, a safety engineer conducts braking tests for 35models of Make A and 35 models of Make B. The mean braking distance for Make A is 41feet. Assume the population standard deviation is 4.6 feet.The mean braking distance for Make B is 45feet. Assume the population standard deviation is 4.3 feet.At alphaequals0.10​,can the engineer support the claim that the mean braking distances are different for the two makes of​ automobiles

Sagot :

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.