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It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportation researcher wants to determine if the statement made in the advertisement is false. She randomly test drives 37 small cars at 65 miles per hour and records the braking distance. The sample average braking distance is computed as 111 feet. Assume that the population standard deviation is 21 feet.

Required:
a. State the null and the alternative hypotheses for the test.
b. Calculate the value of the test statistic.
c. Find the p-value. 0.01 p-value.
d. Use α = 0.01 to determine if the average breaking distance differs from 120 feet.

Respuesta :

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Answer:

Step-by-step explanation:

Given that :

population mean μ = 120

sample size n = 37

sample mean x = 111

standard deviation = 21

The null hypothesis and the alternative hypothesis can be computed as:

The null hypothesis:

[tex]\mathbf{ H_o: \mu = 120 }[/tex]

The alternative hypothesis:

[tex]\mathbf{ H_o: \mu \neq 120 }[/tex]

Since this test is two-tailed,  the value for the test statistics can be computed as:

[tex]z = \dfrac{\overline x - \mu}{\dfrac{\sigma }{\sqrt{n}}}[/tex]

[tex]z = \dfrac{111 - 120}{\dfrac{21}{\sqrt{37}}}[/tex]

[tex]z = \dfrac{-9}{\dfrac{21}{6.08}}[/tex]

z = -2.605

z = -2.61

Since this is a two-tailed test

P-value = 2(z < -2.61)

From the z table;

P-value = 2(0.0045)

P -value = 0.009

Decision rule: To reject the null hypothesis if the p-value is lesser than the level of significance

Conclusion: We reject the null hypothesis since the p-value is lesser than the level of significance. Thus, there is sufficient evidence to conclude that the average breaking distance differs from 120 feet.

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