Question

You draw a simple random sample of 100 taxi drivers living in New York City. You wish to test whether mean time spent driving for New York City taxi drivers is equal to 7 hours. You obtain a test statistic of 2.4. Using a 1% significance level, what is the outcome of this test

Answer 1

Answer:

$t = 2.4$

The degrees of freedom are given by:

$df=n-1=100-1=99$  

And the p value would be given by:

$p_v =2*P(t_{(99)}>2.4)=0.0183$  

For this case we see that the p value is higher than the significance level so we don't have enough evidence to reject the null hypothesis and then the true mean is not significantly different from 7 at the significance level given

Step-by-step explanation:

Data given and notation  

$s$ represent the sample standard deviation for the sample  

$n=100$ sample size

$\mu_o =7$ represent the value to test

$\alpha=0.01$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is iequal to 7 or no, the system of hypothesis would be:  

Null hypothesis:$\mu = 7$  

Alternative hypothesis:$\mu \neq 7$  

The statitic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

For this case we know that the statistic is:

$t = 2.4$

The degrees of freedom are given by:

$df=n-1=100-1=99$  

And the p value would be given by:

$p_v =2*P(t_{(99)}>2.4)=0.0183$  

For this case we see that the p value is higher than the significance level so we don't have enough evidence to reject the null hypothesis and then the true mean is not significantly different from 7 at the significance level given