A company produces an item M whose mean sales per shop per month was found to be 84 dozens. After an advertisement, a sample of 26 shops was taken and mean sales was found to be 80 dozens with standard deviation of 9. Based on this data, which of the following is true?
(Given \( \mathrm{t_{25}\,(0.05)=2.06} \) )
[t = t-test statistic]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Advertisement was effective for sales
The test checks whether the mean sales after the advertisement differ significantly from 84 dozens.
\( H_0 \): μ = 84 (the advertisement has made no difference)
\( H_1 \): μ ≠ 84
Here n = 26, \( \bar x \) = 80 and the sample standard deviation s = 9, so
\( t = \displaystyle\frac{\bar x - \mu}{s/\sqrt{n-1}} = \displaystyle\frac{80-84}{9/5} = -2.22 \)
\( |t| = 2.22 > t_{25}(0.05) = 2.06 \)
So \( H_0 \) is rejected at the 5% level. The change in mean sales after the advertisement is significant, that is, the advertisement had a significant effect on sales.
Why the other options are incorrect
Option (1) Advertisement was not effective for sales — Incorrect. That would mean the change is not significant, which needs |t| < 2.06.
Option (3) Calculated | t | = \( \mathrm{t_{25}\,(0.05)} \) — Incorrect. Calculated |t| = 2.22, not 2.06.
Option (4) Calculated | t | < \( \mathrm{t_{25}\,(0.05)} \) — Incorrect. Calculated |t| = 2.22 is more than 2.06.