Question

An entrepreneur is considering the purchase of a coin operated laundry. The current owner claims that over the past 5 years, the mean daily revenue(u) was $600 with a population standard deviation (a ) of $150. A sample of 30 days (n) reveals a daily mean revenue of $750 (sample mean, X bar), mean revenue is not equal to $600 and assume that the level of significance(a -005) please answer the following questions (20 points) If you were to test the altemate hypothesis that the daily 1. State your null and alternate hypotheses: 2 What is the value of test statistic? Please show all the relevant calculations 3 What is the rejection criterion based on critical value approach? 4. What is the Statistical decision (L.e. reject /or do not reject the null hypothesis)? Provide justification for your decision.

Answer 1

Answer

(1) Claim is given as "daily mean revenue is equal to $600". So, we have to test the alternate hypothesis "daily mean revenue is not equal to $600" (this is given in the question)

So, null hypothesis will be

And alternate hypothesis will be

(2) Population standard deviation is known, so we will use z distribution for the calculation.

Formula for the z statistics is given as

where we have

x(bar) is sample mean, mu is population mean, n is sample size and sigma is population standard deviation.

setting the given values

(3) Using z distribution table for 0.05 significance level, we get z critical value = -1.96 and +1.96

So, we will reject claim if calculated z statistic falls within range of -1.96 and 1.96

(4) It is clear that the z statistics is 5.48 which is falling outside the range of -1.96 and 1.96. This means we can reject the null hypothesis because there is enough evidence to support the alternate hypothesis or we can say that we can reject the claim "daily mean revenue is equal to $600". Result is significant.