Respuesta :
Answer:
[tex]t=\frac{28.9-30}{\frac{19}{\sqrt{240}}}=-0.897[/tex]
[tex]p_v =P(t_{239}<-0.897)=0.185[/tex]
If we compare the p value with a significance level for given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean is lower than 30 minutes or half of and hour at 5% of signficance.
Step-by-step explanation:
Previous concepts
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
For this case since the sample size is large enough n = 240>30 we can assume that the distribution for the sample mean is normal.
Data given and notation
[tex]\bar X=28.9[/tex] represent the sample mean
[tex]s=19[/tex] represent the sample standard deviation for the sample
[tex]n=240[/tex] sample size
[tex]\mu_o =30[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean conmute time in U.S is less than 30 min, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 30[/tex]
Alternative hypothesis:[tex]\mu < 30[/tex]
We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{28.9-30}{\frac{19}{\sqrt{240}}}=-0.897[/tex]
Calculate the P-value
The degrees of freedom are given by:
[tex] df= n-1= 240-1= 239[/tex]
Since is a one-side lower test the p value would be:
[tex]p_v =P(t_{239}<-0.897)=0.185[/tex]
Conclusion
If we compare the p value with a significance level for given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean is lower than 30 minutes or half of and hour at 5% of signficance.
