Question

ACME Fruit Distributing knows that for the last four years, the mean (or average) time it takes a particular truck to make its rounds has been 6.32 hours with standard deviation 0.41hours.

(a) Suppose you plan to observe this truck 40times. What is the approximate distribution of the random sample mean time? Also specify its parameters.

(b)What is the approximate probability that this random sample mean (with n=40) will be more than 6.7 hours?

(c) The truck route driver recently retired and Kyle took his place. Would observing Kyle's time on one particular day to be 6.1 hours cause you to think he is certainly a better driver than the previous one? Explain.

(d) Would observing Kyle's mean(average)time on 40 consecutive days to be 6.1 hours cause you to think he is certainly a better driver than the previous one? Explain.

(a), (b) are okay!

I can't find the proper answer. Unfortunately, I haven't learned the p-value. Also, I can't understand the meaning of (c), (d) questions.

Only one particular 6.1 hours can be determined he is better than previous one?

Answer 1

Answer:-

Given that:-

ACME Fruit Distributing knows that for the last four years, the mean (or average) time it takes a particular truck to make its rounds has been 6.32 hours with standard deviation 0.41hours.

As n is quite large to conclude its distribution to be normal.

According to Central limit theorem The distribution of sample mean is normal with mean 6.32 and standard deviation

If the random variable X denotes the time to make rounds

b)

P(Z > 5.86) = 0 ( probability value is obtained from standard normal table)

c)

The time noted on the very first day of his driving is 6.1 hours. Based on this one day information we can conclude undoubtedly that kyle is a better driver than the old one. Since no other information on his driving is available. we are left with only one observation not a data.

d)

Here we have a data on his driving for 40 days and mean is observed to be 6.1. Inorder to conduct a two sample test to test for the difference between means of the driving time we need to have standard deviation of kyle's driving time which is not available. Hence we do a one sample z test and will test the hypothesis

Test statistic,

If tested for 5% significance level of this one tailed test, the table value obtained from standard normal table is -1.645 >z

If tested for 1% significance level of this one tailed test, the table value obtained from standard normal table is -2.326 >z.

So at both the significance level the calculated value is less than table value, so reject the null hypothesis even at 1% level of significance implies that the kyle is a better driver than old driver in driving time