CUET · MATHS · PYQ PAPER 2025
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 \(\left.t_{16}(0.05)=2.12\right]\)
- A Strategy was effective as null hypothesis is accepted
- B Strategy was effective as null hypothesis is rejected
- C Strategy was not effective as null hypothesis is accepted
- D Strategy was not effective as null hypothesis is rejected
Answer & Solution
Correct Answer
(B) Strategy was effective as null hypothesis is rejected
Step-by-step Solution
Detailed explanation
\(H_0: \mu = 150\) \(H_1: \mu > 150\) \(t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\) \(t = \frac{165 - 150}{25/\sqrt{17}}\) \(t = \frac{15}{25/4.123}\) \(t = \frac{15}{6.063}\) \(t \approx 2.474\) Since \(t_{calculated} (2.474) > t_{critical} (2.12)\), reject \(H_0\). Strategy was…
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