SOLUTION: 9.47
The waiting time to place an order at a branch of a fast-food chain during the lunch hour has had a population mean of 3.55 minutes, with a population standard deviation of
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Question 358596: 9.47
The waiting time to place an order at a branch of a fast-food chain during the lunch hour has had a population mean of 3.55 minutes, with a population standard deviation of 1.1 minutes. Recently, in an effort to reduce the waiting time, the branch has experimented with a system in which there is a single waiting line. A sample of 100 customers during a recent lunch hour was selected, and their mean waiting time to place an order was 3.18 minutes. Assume that that the population standard deviation of the waiting time has not changed from 1.1 minutes.
a. At the level of significance, using the critical value approach to the hypothesis testing, is there evidence that the population mean waiting time to place an order is less than 3.55 minutes?
b. At the level of significance, using the p-value approach to the hypothesis testing, is there evidence that the population mean waiting time to place an order is less than 3.55 minutes?
c. Interpret the meaning of the p-value in this position.
d. Compare your conclusions in a and b.
Answer by jrfrunner(365) (Show Source): You can put this solution on YOUR website!
Given
and
sample n=100 and sample mean xbar=3.18
--
hypothesis
Ho:
Ha: lower tail test
Note: I dont see where you provide the signifance level to perform the test
I'll assume
==
a. At the level of significance, using the critical value approach to the hypothesis testing, is there evidence that the population mean waiting time to place an order is less than 3.55 minutes?
we dont know the population distribution. But since we are testing sample averages and the sample size is >30 we can rely on the "Central Limit Theorem" to assume a normal distribution for the sample averages.
--
furthermore since we are told the population standard deviation
we know that we can use a critical value based on Z
--
lower tail critical value is Z at a tail probability of
Z=-1.645
What this means is that sample standardized values as extreme as -1.645 can be expected by pure chance if the null hypothesis is assumed
--
Compute the test statistic or Z=(3.18-3.55)/(1.1/sqrt(100))=-3.36
Since our test statistic (our evidence for the null) is more extreme than the critical value of Z=-1.645 then we reject the null. This cannot happen by pure chance.
==
b. At the level of significance, using the p-value approach to the hypothesis testing, is there evidence that the population mean waiting time to place an order is less than 3.55 minutes?
pvalue =P(Xbar<3.18) given the assumption of the null
P(Xbar<3.18)=P(Z<-3.36)=very small (less than 0.005) So we conclude that a value of Xbar as extreme as 3.18 cannot happen by pure chance under the null, so we reject the null hypothesis and conclude that evidence suggest
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