Question

A recent study claims that business travelers spend an average of $39 per day on meals. A sample of 15 business travelers found that they had spent an average of $42 per day with a standard deviation of $3.78. If alpha =0.05, what is the claim, should we reject H0 or fail to reject H₀, what is the test value, what is the alternate hypothesis?

Answer 1

Solution:

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 business travelers spend an average of $39 per day on meals. (Claim)

Alternative hypothesis: Ha: The business travelers spend an average different than $39 per day on meals.

H0: µ = 39 versus Ha: µ ≠ 39

This is a two tailed test.

Claim is the null hypothesis.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 39

Xbar = 42

S = 3.78

n = 15

df = n – 1 = 14

α = 0.05

Critical value = - 2.1448 and 2.1448

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (42 - 39)/[3.78/sqrt(15)]

t = 3.0738

P-value = 0.0083

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the business travelers spend an average of $39 per day on meals.