Question

"A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.3 minutes with a standard deviation of 1.5 minutes. At the 0.01 significance level, test the claim that the mean waiting time is less than 10 minutes. Use the P-value method of testing hypotheses."

Answer 1

We conclude that the mean waiting time is less than 10 minutes.

Explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.3 minutes with a standard deviation of 1.5 minutes.

Let
`\mu` = mean waiting time.

So, Null Hypothesis,
`H_0` :
`\mu`
`\geq` 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis,
`H_A` :
`\mu` < 10 minutes {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. =
`(\bar X-\mu)/((s)/(√(n)))` ~
`t_n_-_1`

where,
`\bar X` = sample mean waiting time = 7.3 minutes

s = sample standard deviation = 1.5 minutes

n = sample of occasions = 18

So, test statistics =
`(7.3-10)/((1.5)/(√(18)))` ~
`t_1_7`

= -7.637

The value of t test statistics is -7.637.

Now, the P-value of the test statistics is given by the following formula;

P-value = P(
`t_1_7` < -7.637) = Less than 0.05%

Since, our P-values of test statistics is less than the level of significance as 0.05% < 1%, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the mean waiting time is less than 10 minutes.