Question

A 2015 study by Zagat found that Americans pay an average of $3.28 for a cup of coffee. I have a hunch that the average price for a cup of coffee my online students pay will be different, but I am not sure if it will be higher or lower than the national average. So, I asked 16 of my students to report how much they typically pay for a cup of coffee. I assume that coffee prices in the Zagat study are normally distributed, and I set the significance level at α = .05.

Student	Coffee Cup Price
1	2.23
2	4.01
3	3.03
4	1.79
5	2.44
6	3.18
7	4.01
8	3.12
9	1.25
10	2.79
11	4.25
12	3.56
13	3.34
14	1.65
15	3.76
16	2.49

1. What is the dependent variable in this hypothesis test? What is the “sample” for this one-sample t test?
2. I am interested in testing whether my students pay different prices for their coffee than the national average, but I am not sure if my students pay more or less than the national average. That means my hypothesis is non-directional. What would be the null and alternative hypotheses in both words and symbol notations for my analysis?
3. Calculate the sample mean.
4. Estimate the standard deviation of the population.
5. Calculate the standard error (standard deviation of the sampling distribution).
6. Calculate the tstatistic for the sample. Specify whether the test is a one-tailed or two-tailed test based on the hypotheses form (b), and then determine the critical tvalue(s) based on the type of test and the preset alpha level.

7. Compare thetstatistic with the critical t value. Is the calculated t statistic more extreme or less extreme than the critical t value? Then make a decision by stating whether we “reject” or “fail to reject” the null hypothesis.  Interpret the result in 1-2 sentences (you may restate the hypothesis accepted or explain it in your own words).

8. Calculate the raw effect size and the standardized effect size of this hypothesis test.

Answer 1

(a)

The dependent variable in this hypothesis test : Coffee Cup Price

The “sample” for this one-sample t test : Coffee Cup Price of 16 Students

(b)

H0: Null Hypothesis: $\mu$ = $3.28 ( The average price for a cup of coffee my online students pay will not be different from the national average )

HA: Alternative Hypothesis: $\mu$$\neq$ $3.28 ( The average price for a cup of coffee my online students pay will be different from the national average ) (claim)

(c)

Sample mean = $\bar{x}$ = 46.9/16 = 2.9313

(d)

Estimate the standard deviation of the population = s =

12.1414 15 = V0.8094 = 0.8997 V

(e)

The standard error (standard deviation of the sampling distribution) = s/$\sqrt{n}$

                                                                                                = 0.8997/$\sqrt{16}$
                                                                                       = 0.2249

(f)

Test Statistic is given by:

t = (2.9313 - 3.28)/0.2249

= - 1.5505

The test is a two-tailed test

$\alpha$ = 0.05

ndf = 16 - 1 = 15

From Table, critical values of t = $\pm$2.1314

(g)
Since calculated value of t = - 1.5505 is greater than critical value of t = - 2.1314, the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data do not support the claim that the average price for a cup of coffee my online students pay will be different from the national average

Raw effect size = 3.28 - 2.9313 = 0.3487

Standardized effect size = 0.3487/0.8997

                                  = 0.3876