Respuesta :
Answer:
We conclude that the mean waiting time is less than 10 minutes.
Step-by-step 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 7.8 minutes with a standard deviation of 2.5 minutes.
Let [tex]\mu[/tex] = mean waiting time for bus number 14.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 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 the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean waiting time = 7.8 minutes
s = sample standard deviation = 2.5 minutes
n = sample of different occasions = 18
So, test statistics = [tex]\frac{7.8-10}{\frac{2.5}{\sqrt{18} } }[/tex] ~ [tex]t_1_7[/tex]
= -3.734
The value of t test statistics is -3.734.
Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.
Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the mean waiting time is less than 10 minutes.
Using the t-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, it is found that there is enough evidence to support the claim that the mean is less than 10 minutes.
At the null hypothesis, it is tested if the mean is of at least 10 minutes, that is:
[tex]H_0: \mu \geq 10[/tex]
At the alternative hypothesis, it is tested if it is less than 10 minutes, that is:
[tex]H_1: \mu < 10[/tex]
We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
- n is the sample size.
For this problem, the values of the parameters are: [tex]\overline{x} = 7.8, \mu = 10, s = 2.5, n = 18[/tex]
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{7.8 - 10}{\frac{2.5}{\sqrt{18}}}[/tex]
[tex]t = -3.73[/tex]
The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a 0.01 significance level and 18 - 1 = 17 df is of [tex]t^{\ast} = -2.57[/tex]
Since the test statistic is less than the critical value for the left-tailed test, it is found that there is enough evidence to support the claim that the mean is less than 10 minutes.
A similar problem is given at https://brainly.com/question/13873630
