A small start-up started making wafers and distributing them to the retailers. After a week, average sales per week were found to be 150 packets. So, to increase the sales, a strategy was used to change the packaging and add a chocolate worth Rs. 5 as a free gift with the pack. After this, a sample of 17 shops was taken, which showed that sales went up with mean 165 and a standard deviation of 25. Check whether the strategy was effective @5%, level of significance.? [Given \(t_{16}(0.05)=2.12\)]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Strategy was effective as null hypothesis is rejected
Let \(\mu\) be the mean weekly sales after the new strategy. The null hypothesis says the strategy made no difference:
\( H_0: \mu = 150 \)
\( H_1: \mu \gt 150 \)
The sample is small (n = 17) and the population standard deviation is unknown, so the t-test is used with \(n - 1 = 16\) degrees of freedom. Taking s = 25 as the standard deviation of the sample (found with divisor n), the standard error is \(\frac{s}{\sqrt{n-1}}\):
\( \displaystyle t = \frac{\bar{x} - \mu}{s/\sqrt{n-1}} \)
\( \displaystyle = \frac{165 - 150}{25/\sqrt{16}} \)
\( \displaystyle = \frac{15}{6.25} = 2.4 \)
Since \(2.4 \gt 2.12\), the calculated value exceeds the table value, so \(H_0\) is rejected: the sales have risen significantly, and the strategy was effective. (Taking the standard error as \(\frac{s}{\sqrt{n}}\) instead gives \(t \approx 2.47\), and the conclusion is the same.)