Question

An online education company is considering selling a new online course for $450. They want to know if their existing customers would be willing to pay that amount, so they randomly sample 20 of their existing customers and ask them what they would be willing to pay. Among the responses, the average is $420, and the SD is $60. A histogram of the samples approximately follows the normal curve. In this problem we will conduct a t-test for the null hypothesis that the average response among all existing customers is $450, versus the alternative hypothesis that the average response among all existing customers is less than this.

(a) Under the null hypothesis, the sample average is expected to be $___________. The 
standard error is estimated to be $___________. 
 
 
 


(b) The t-test statistic is __________. 


(c) The p-value is ___________________________________________. (Give a range of values.) 


(d) Our conclusion is (circle one) reject the null hypothesis OR don’t reject the null hypothesis.

Answer 1

(a) Under the null hypothesis, the sample average is expected to be $450. The 
standard error is estimated to be $13.4164.

(b) The t-test statistic is -2.2361 


(c) The p-value is 0.0188.

(d) Our conclusion is to reject the null hypothesis.

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 average response among all existing customers is $450

Alternative hypothesis: Ha: the average response among all existing customers is less than $450

H0: µ = 450 versus Ha: µ < 450

This is a lower tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 450

Xbar = 420

S = 60

n = 20

df = n – 1 = 19

α = 0.05

Critical value = -1.7291

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (420 - 450)/[60/sqrt(20)]

t = -2.2361

P-value = 0.0188

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the average response among all existing customers is less than $450