“The See Me” marketing agency wants to determine if time of day for a television advertisement influences website searches for a product. They have extracted the number of website searches occurring during a one-hour period after an advertisement was aired for a random sample of 30 day and 30 evening advertisements. A portion of the data is shown in the accompanying table. |
Day Searches | Evening Searches |
96670 | 118379 |
97855 | 111005 |
96368 | 100482 |
98465 | 122160 |
98550 | 117158 |
101623 | 101556 |
95753 | 98875 |
102036 | 104384 |
99475 | 110932 |
103780 | 101963 |
97608 | 123513 |
99859 | 102195 |
101764 | 111388 |
97287 | 116287 |
98066 | 119660 |
95390 | 112553 |
96125 | 98245 |
96767 | 98062 |
99494 | 101657 |
102498 | 106451 |
99260 | 104247 |
102020 | 105507 |
102468 | 118339 |
96543 | 109847 |
102491 | 123996 |
96557 | 122545 |
102627 | 113248 |
95048 | 104941 |
96969 | 111829 |
95103 | 114721 |
a. |
Set up the hypotheses to test whether the mean number of website searches differs between the day and evening advertisements. |
a) H_{0}: μ_{1} − μ_{2} = 0; H_{A}: μ_{1} − μ_{2} ≠ 0 b) H_{0}: μ_{1} − μ_{2} ≥ 0; H_{A}: μ_{1} − μ_{2} < 0 c) H_{0}: μ_{1} − μ_{2} ≤ 0; H_{A}: μ_{1} − μ_{2} > 0 |
b-1. |
Find the value of the test statistic. Assume the population variances are unknown but equal. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.) |
Test statistic = |
b-2. |
Assume the population variances are unknown but equal. What are the critical value(s) and the rejection rule? (Negative value should be indicated by a minus sign. Round your answers to 4 decimal places.) |
Reject H_{0} if t_{df}> ? or t_{df}< ? . |
c. |
At the 5% significance level, what is the conclusion? |
(Click to select) Reject or Do not reject H_{0}.At the 5% significance level, we conclude the mean number of website searches (Click to select) differs or not differ between the day and evening advertisements. |
a)
a) H_{0}: μ_{1} − μ_{2} = 0; H_{A}: μ_{1} − μ_{2} ≠ 0
b)
value of the test statistic t =-7.249
b-2)
Reject H_{0} if t >2.002 to t<-2.002
c)
Reject H_{0}.At the 5% significance level, we conclude the mean number of website searches differs between the day and evening advertisements.
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