Set up the null and alternative hypotheses and discuss the consequences of Type I error and Type II error.
The manager of a department store is thinking about establishing a new billing system for the store's credit customers. After a thorough financial analysis, she determines that the new system will be cost-effective only if the mean monthly account is more than $170. A random sample of 400 monthly accounts is drawn, for which the sample mean is $178. The manager knows that the accounts are approximately normally distributed with a standard deviation of $65. Can the manager conclude that the new system would not be cost effective?
1 Answer | Add Yours
We wish to test the null vs the alternative hypotheses `H_0`, `H_1` resepctively, where
`H_0 : mu_0 = 170`
`H_1: mu_1 >=170`
This is a one-sided hypothesis test.
Given the sample size 400 and sample mean 178 and that the monthly accounts are approximately Normal with mean `mu` and standard deviation 65, the distribution of the sample mean is Normal(`mu`, 65/sqrt(400)).
Therefore, the required test statistic is
`z = (178- mu_0)/(65/20) = 2.46`
The 95th percentage point of the standard Normal distribution (which z follows) is 1.645, so the manager can reject the null at the 5% level of significance and conclude that the new system may be cost effective.
Type I error: Pr(reject H0| H0 true) = 0.05
Type II error: Pr(not reject H0| H1 true) - Supposing the true mean is 175 say
= `Phi((170-175)/(65/20)) = Phi(-1.538) = 0.06`
We fix the chance of concluding the system is cost effective when it isn't at 5%. If the true monthly mean is $175, the chance of concluding the new system is not cost effective (monthly mean = $170) is 6%.
We’ve answered 319,200 questions. We can answer yours, too.Ask a question