Confidence Interval exercises CORRECTED PDF

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A railway company wants to launch a new discount card allowing customers to have special discounts on second-class tickets for people aged between 18 and 25 years old. However, they need to find the best pricing strategy before launching the new product on the market. The company decides to run a survey across the potential customers in order to find the price-range having 90% of chances, containing the price expected by the clients. By randomly selecting a sample of size n=400 young customers, they found that the sample mean is 𝑥 = 250 𝐸𝑢𝑟𝑜 and the variance is equal to 7288. Compute the confidence interval at 90% for the expected card price (population average).

𝑝 (𝑥 − 𝑧

𝜎

√𝑛

< 𝜇 < 𝑥 + 𝑧

𝜎

√𝑛

) = 0.9 = 𝑝 (250 − 1.64 (

85.37 85.37 ) < 𝜇 < 250 + 1.64 ( )) = 20 20

𝑝(243 < 𝜇 < 257) = 0.9 Assuming that 10000 clients will buy the card, can you identify the confidence interval of the revenue the company will make? 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑝𝑟𝑖𝑐𝑒 × 10000. 𝑇ℎ𝑒𝑛 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑝 (10000 ∗ 243 < 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 < 10000 ∗ 257) = 0.9

Therefore 𝑝(2430000 𝐸𝑢𝑟𝑜 < 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 < 2570000 𝐸𝑢𝑟𝑜) = 0.9

A supermarket wants to launch a new service, delivering food at home for free, only for those who buy food online. However, the service will be launched only if the company will be sure, at 95% level, that more than 30% of their customers will use it. 400 customers were interviewed and only 100 claimed they will use the service. What type of conclusions we can provide to the supermarket?

𝑝 (𝑝 − 𝑧 𝑝 (0.25 − 1.96

√𝑝(1 − 𝑝) √𝑛

√0.25(1 − 0.25) √400

< 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 < 𝑝 + 𝑧

√𝑝(1 − 𝑝) ) = 0.95 √𝑛

< 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 < 0.25 + 1.96

√0.25(1 − 0.25)

𝑝(0.208 < 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 < 0.292) = 0.95

√400

) = 0.95

The interval does not include 30% and is indeed clearly below it so…We suggest the supermarket not to launch this service since otherwise they will be disappointed about the low use of it made by clients...