Question
Use this information for questions 1-5: According to a 2020 Car and Driver report, the mean age of cars on the road in the US
Use this information for questions 1-5:
According to a 2020 Car and Driver report, the mean age of cars on the road in the US is 11.9 years. Some makers advertise that their cars are more durable. You've noticed older Volvos on the road, and think the average age of Volvos still on the road might be older than 11.9. You examine records of 300 randomly selected Volvo owners and find the average age to be 12.3 years with a standard deviation of 4.7 years.
1.)What is the alternate hypothesis?
a.)Ha: > 12.3
b.)Ha:xx > 11.9
c..)Ha:xx> 12.3
d.)Ha: > 11.9
2.)The age of cars on the road is right-skewed, as is your sample of Volvos, however it has no extreme outliers. Is it acceptable to to use the t-distribution procedures?
a.)No, the sample size is too small.
b.)Yes, the sample of 300 cars is large so that the sample means distribution will be approximately Normal.
c.)No, since the sample is skewed, we can't proceed to do inference.
d.)Yes, skewness is never a problem.
3.)Can we just note that 12.3 is greater than 11.9 and infer that the average age of Volvos is older than 11.9?
a).Yes, it is obvious that Volvo's average age is older and we only perform a hypothesis test to determine if the difference in 12.3 and 11.9 is important.
b.)No, we need to perform a hypothesis test to determine if our sample results could be due to chance alone rather than due to Volvos having a larger mean.
c.)No, we have to perform a significance test to find out which hypothesis is true.
d.)Yes, we just perform a hypothesis test to say that our evidence is good.
4.)Perform the test to determine if the average age of Volvo automobiles still on the road is more than 11.9. Enter the t-test statistic rounded to two decimal places.
Correct format: 2.19
5.)Based on the p-value of the test, you conclude that the evidence is significant at the alpha level of ___.
Mark all that apply.
a.)0.075
b.)0.01
c.)0.05
d.)0.10
e.)0.001
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