Question

A travel agent states that the mean meal price for a family of four in a resort town is at most $100. You decide to test this claim and you take a random sample of meal prices in the resort town. You find that for 33 meals for family of four the mean price is $110 with a standard deviation of $21. Is there enough evidence to reject the travel agent's claim? (a=.05)

what test will you use? T test

state the null hypothesis (Ho)?

state the alternative hypothesis (Ha).

what is the critical probability value (a)?

what is the P value or P calculated?

do you reject null hypothesis or fail to reject null hypothesis?

state your findings?

Answer 1

One sample t-test

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H₀: the mean meal price for a family of four in a resort town is at most $100.

Alternative hypothesis: H_a: the mean meal price for a family of four in a resort town is less than $100.

H₀: µ ≤ 100 versus H_a: µ > 100

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 100

Xbar = 110

S = 21

n = 33

df = n – 1 = 32

α = 0.05

Critical value = 1.6939

(by using t-table or excel)

t = (110 – 100)/[21/sqrt(33)]

t = 2.7355

P-value = 0.0050

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to reject the travel agent's claim.

There is sufficient evidence to conclude that the mean meal price for a family of four in a resort town is less than $100.

There is not sufficient evidence to conclude that the mean meal price for a family of four in a resort town is at most $100.