The Metropolitan Bus Company claims that the mean waiting time for a bus during rush hour is less than 5 minutes. A random sample of 20 waiting times has a mean of 3.7 minutes with a standard deviation of 2.1 minutes. At α = 0.01, test the bus company’s claim. Assume the distribution is normally distributed. State the sample mean and the standard deviation

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Answer:

Decision is reject the [tex]H_0[/tex] as There is sufficient evidence to support the claim that the mean time is less then 5 minutes.

Step-by-step explanation:

Consider the provided information.

The mean waiting time for a bus during rush hour is less than 5 minutes.

Thus the null hypothesis is:

[tex]H_0: \mu\geq 5[/tex]

[tex]H_1: \mu< 5[/tex]

A random sample of 20 waiting times has a mean of 3.7 minutes with a standard deviation of 2.1 minutes. At α = 0.01,

Now use the table to find the z value, where α = 0.01,

[tex]t_L = -2.3264[/tex]

The value of n is 20, sample mean is 3.7, population mean is 5 and standard deviation is 2.1.

Now, use the formula: Standardized test statistic for z-scores =[tex]t=\frac{\bar x-\mu_0}{{s}/{\sqrt{n} }}[/tex].

Substitute the respective values in the above formula.

[tex]t=\frac{3.7-5}{{2.1}/{\sqrt{20} }}[/tex]

[tex]t=\frac{-1.3}{0.47}[/tex]

[tex]t=-2.768465115\approx-2.77[/tex]

As, the test statistic is in the reject interval, so reject [tex]H_0[/tex]

There is sufficient evidence to support the claim that the mean time is less then 5 minutes.

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