Question

A metropolitan transportation authority has set a bus mechanical reliability goal of 3 comma 9003,900 bus miles. Bus mechanical reliability is measured specifically as the number of bus miles between mechanical road calls. Suppose a sample of 100100 buses resulted in a sample mean of 3 comma 9753,975 bus miles and a sample standard deviation of 325325 bus miles. Complete parts​ (a) and​ (b) below. a. Is there evidence that the population mean bus miles is more than 3 comma 9003,900 bus​ miles? (Use a 0.050.05 level of​ significance.)

Answer 1

`t=(3975-3900)/((325)/(√(100)))=2.308`

The degrees of freedom are given by:

`df=n-1=100-1=99`

The p value for this case would be given by:

`p_v =P(t_((99))>2.308)=0.0115`

Since the p value is lower than the significance level we can reject the null hypothesis and then we can conclude that the true mean is significantly higher than 3900 bus miles

Explanation:

Data given

`\bar X=3975` represent the sample mean for the miles

`s=325` represent the sample standard deviation

`n=100` sample size

`\mu_o =3900` represent the value to verify

`\alpha=0.05` represent the significance level

t would represent the statistic

`p_v` represent the p value

System of hypothesis

For this case we want to check if there evidence that the population mean bus miles is more than 3,900 bus​ miles , the system of hypothesis are:

Null hypothesis:
`\mu \leq 3900`

Alternative hypothesis:
`\mu > 3900`

The statistic for this case is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the data given we got:

`t=(3975-3900)/((325)/(√(100)))=2.308`

The degrees of freedom are given by:

`df=n-1=100-1=99`

The p value for this case would be given by:

`p_v =P(t_((99))>2.308)=0.0115`

Since the p value is lower than the significance level we can reject the null hypothesis and then we can conclude that the true mean is significantly higher than 3900 bus miles