In: Economics
To better to assess your willingness-to-pay for advertising on others’ websites, you want to learn the mean profit per visit for all visits to your website. To accomplish this, you have collected a random sample of 50 visits to your website over the past 6 months. This sample includes information on visit duration and profits.
Let the null hypothesis be the mean profit per visit for all of your visitors be $11.50. (a) Calculate the corresponding t-stat for this null hypothesis. (b) Calculate the corresponding p-value for this null hypothesis. (c)With strength of 95% decide whether or not to reject this null hypothesis. (d) Detail the reasoning behind your decision.
**Include excel details, if possible.**
Visitor | Duration | Profit |
1 | 2 | 0 |
2 | 7 | 15 |
3 | 5 | 0 |
4 | 8 | 0 |
5 | 16 | 0 |
6 | 8 | 0 |
7 | 9 | 0 |
8 | 6 | 0 |
9 | 5 | 0 |
10 | 28 | 0 |
11 | 3 | 0 |
12 | 5 | 18 |
13 | 12 | 0 |
14 | 10 | 16 |
15 | 10 | 0 |
16 | 15 | 0 |
17 | 21 | 0 |
18 | 10 | 46 |
19 | 20 | 21 |
20 | 4 | 0 |
21 | 10 | 15 |
22 | 9 | 0 |
23 | 6 | 22 |
24 | 8 | 0 |
25 | 6 | 0 |
26 | 8 | 17 |
27 | 5 | 0 |
28 | 2 | 0 |
29 | 14 | 28 |
30 | 10 | 0 |
31 | 7 | 0 |
32 | 12 | 0 |
33 | 3 | 0 |
34 | 7 | 21 |
35 | 9 | 0 |
36 | 4 | 9 |
37 | 5 | 27 |
38 | 12 | 0 |
39 | 11 | 21 |
40 | 5 | 0 |
41 | 5 | 0 |
42 | 10 | 0 |
43 | 21 | 0 |
44 | 7 | 0 |
45 | 5 | 0 |
46 | 6 | 18 |
47 | 7 | 0 |
48 | 4 | 21 |
49 | 4 | 33 |
50 | 11 | 0 |
Let the null hypothesis be H0: Profit = 11.5 and alternative hypothesis be H1: Profit < 11.5 (one-tailed test). We may also go for two tailed test, but let us consider one-tailed, specifically left tailed for this case, so that we will be able to test whether the sample mean profit is less than the hypothesized value or equal to it.
(a) The t-statistic for the corresponding test can be calculated as t = (Sample mean - hypothesized mean) / (standard error), where sample mean = 6.96, hypothesized mean = 11.5, standard error = sample standard deviation / sqrt(number of observations) = 11.31/sqrt(50) = 1.6. Hence, t-stat = (6.96 - 11.5)/1.6 = -2.84
(b) The p-value corresponding to the t-statistic = 0.003.
(c) Since p-value is < 0.05 or (1 - 0.95), we have strong statistical evidence in favor of rejecting the null hypothesis.
(d) Since we reject the null hypothesis based on the p-value (refer c), we conclude that average profit < $11.5
The calculated values are given in the left side table, whereas the formula view of the calculations is given on the right hand side.