Question

# A bus company advertised a mean time of 150 minutes for a trip between two cities....

A bus company advertised a mean time of 150 minutes for a trip between two cities. A consumer group had reason to believe that the mean time was significantly different from 150 minutes. A sample of 40 trips showed a mean x̄= 153 minutes and a standard deviation s=7.5 minutes. At the 4% level of significance, test the advertisement claim. Apply two-sided test, and give all six steps of the procedure clearly. Finally interpret the conclusion in terms of the example

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: The mean time for a trip between two cities was 150 minutes.

Alternative hypothesis: Ha: The mean time for a trip between two cities was significantly different from 150 minutes.

H0: µ = 150 versus Ha: µ ≠ 150

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 150

Xbar = 153

S = 7.5

n = 40

df = n – 1 = 39

α = 0.04

Critical value = ±2.1247

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (153 - 150)/[7.5/sqrt(40)]

t = 2.5298

P-value = 0.0156

(by using t-table)

P-value < α = 0.04

So, we reject the null hypothesis

There is sufficient evidence to conclude that the mean time for a trip between two cities was significantly different from 150 minutes.

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