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 $t_{16}(0.05)=2.12$]

Strategy was effective as null hypothesis is rejected

The correct answer is Option (2) → Strategy was effective as null hypothesis is rejected **

Population mean (old average): $\mu_0 = 150$

Sample mean: $\bar{x} = 165$

Sample size: $n = 17$

Sample standard deviation: $s = 25$

Significance level: $5\%$

Hypotheses:

$H_0 : \mu = 150$ (strategy not effective)

$H_1 : \mu > 150$ (strategy effective)

Test statistic:

$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$

$t = \frac{165 - 150}{25 / \sqrt{17}}$

$t = \frac{15}{25 / \sqrt{17}}$

$t = \frac{15}{6.06}$

$t \approx 2.475$

Critical value (given): $t_{16}(0.05) = 2.12$

Since $2.475 > 2.12$, reject $H_0$.

Conclusion: Strategy was effective as null hypothesis is rejected.