Ahh, the holidays-you gotta love 'em. Nothing says holiday quite like eating too much food. Whether it is Thanksgiving, Christmas, or the 4^th of July, we love to eat on holidays. I wonder whether people spend more money at the grocery store two weeks before a major holiday or two weeks after. So I select a random sample of 25 people and see how much they spend on a trip to the store two weeks before the holiday and again two weeks after. Before the holiday, the shoppers spent an average of $80. After the holiday, each shopper spent $85, on average. The standard DEVIATION of the DIFFERENCE between the means was 10.
a. Write the null and alternative hypotheses.
b. Select an alpha level and wrap words around it.
c. Is this a statistically significant difference? (do the calculations)

Question

Quizplus · Accepted Answer

The Answer is a. H₀:

\mu

_x =

\mu

_y
Note that we need to specify here that the X population is shoppers before the holidays and the Y population of shoppers after the holidays. Need to be clear on which population is which.
H₁:

\mu

_x

eq

\mu

_yNote that I used this non-directional alternative hypothesis because the question says they could spend more before OR after the holidays
b. Alpha is .05, meaning that there must be less than a 5% chance that my results are due to chance, or random sampling error, before I will consider them statistically significant.
c. First I need to find the standard error of the differences between the sample means. I'm already given the standard deviation of the difference. So I take that number and divide it by the square root of N, which is the number of pairs in the sample. So:

s_{\bar{D}}

= 10 /

\surd

25

ightarrow

10/5

ightarrow

2
Now I calculate the observed t value:

t=\frac{\bar{X}-\bar{Y}}{s_{\overline{D}}}

ightarrow

t=\frac{80-85}{2}

ightarrow

t=\frac{-5}{2}

ightarrow

t = -2.50. Because this is a 2-tailed test, I can take the absolute value and get an observed t value of 2.50.
Using Appendix B, 2-tailed test, .05 alpha, df = 25 - 1 = 24, I find a critical value of ±2.064.
Because my critical t value is less than my observed t value, I reject the null hypothesis and conclude that these results were not due to chance. In the populations of grocery shoppers, people spend more money on average after the holidays than they do before.

Ahh, the Holidays-You Gotta Love 'Em