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A large university is interested in learning about the average time it takes students to drive to campus. The university sampled 238 students and asked each to provide the amount of time they spent traveling to campus. This variable, travel time, was then used conduct a test of hypothesis. The goal was to determine if the average travel time of all the university's students differed from 20 minutes. Suppose the sample mean and sample standard deviation were calculated to be 23.2 and 20.26 minutes, respectively.

Calculate the value of the test statistic to be used in the test.

A) z = 2.551

B) z = 37.59

C) z = 2.437

D) z = 0.173

Answer :

Answer:

C) z = 2.437

Step-by-step explanation:

The null hypothesis is:

[tex]H_{0} = 20[/tex]

The alternate hypotesis is:

[tex]H_{1} \neq 20[/tex]

Our test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

In this problem, we have that:

[tex]X = 23.1, \mu = 20, \sigma = 20.26, n = 238[/tex]

So

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{23.2 - 20}{\frac{20.26}{\sqrt{238}}}[/tex]

[tex]z = 2.437[/tex]

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