A marketing research company claims that the average cost to have lunch in a res

Question

A marketing research company claims that the average cost to have lunch in a restaurant in Sunshine Coast, Queensland, is less than $30 per person, with a standard deviation of $7. To test this claim, a random sample of 25 persons that had lunch in Sunshine Coast restaurants was taken, and produced a sample mean cost of $27 per person.

(a) Is there evidence at 0.05 level of significance that the mean cost of having lunch in Sunshine Coast restaurants is less than $30 per person?

(b) What assumptions are made in order to conduct the hypothesis test in (a)?

Explanation / Answer

The sample size is 25, which is less than 30, so we will use a t-statistic with 24 degrees of freedom to test the hypotheses.

Null hypothesis : Average cost of lunch is $30

Alternate hypothesis : average cost of lunch is less than $30.

(a) Test statistic = (27-30) / (7/sqrt(25)) = - 3/1.4 = - 2.14286

The critical t-value can be found from a standard t-table for 24 degrees of freedom and one-tailed alpha=0.05 as 1.711

So the rejection region for our test is <-1.711

Since our test statistic - 2.14286 < - 1.711, it lies in the rejection region and hence the Null hypothesis can be rejected.

So we can conclude that the mean cost of having lunch is less than $30.

(b) The general assumptions for t-test are normality of the data distribution, random sampling, scale of measurement and equality of variance.

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