Question

To increase business, the owner of a shoe store is running a promotion in which a customer's bill can be selected at random to receive a discount. When a customer's bill is printed, a program in the cash register determines randomly whether the customer will receive a discount on the bill. The program was written to generate a discount with a probability of 0.15; that is, 15% of the bills get a discount in the long run. However, the owner is concerned the program is incorrect and is not generating the intended long-run proportion of 0.15.

The owner selects a random sample of bills and finds that only 12% of them received a discount. A confidence interval for p, the proportion of bills that will receive a discount in the long run, is 0.12 ± 0.05, and all conditions for inference are met.

Consider the confidence interval 0.12 ± 0.05.

Part A: Does the confidence interval provide convincing statistical evidence that the program is NOT working as intended? Justify your answer. (3 points)

Part B: Does the confidence interval provide convincing statistical evidence that the program generates the discount with a probability of 0.15? Justify your answer. (2 points)

Part C: A second random sample of bills is taken that is six times the size of the original sample. In the second sample, 12% of the bills received the discount. Determine the value of the margin of error based on the second sample of bills used to compute an interval for p with the same confidence level as that of the original interval. (2 points)

Part D: Based on the margin of error in part C that was obtained from the second sample, is the program working as intended? Justify your answer. (3 points) (10 points)

Please answer with a description and work!!!

Answer 1

Solution

Let X = Number of customers who get the discount in a random sample of n customers. Then,

X ~ B(n, p), where p = probability a customer gets the discount.

Back-up Theory

100(1 - α) % Confidence Interval for the population proportion, p is:

p_cap ± Z_α/2[√{p_cap(1 –p_cap)/n}] ………………………………………………………………. (1)

where

Z_α/2 is the upper (α/2)% point of N(0, 1),

pcap = sample proportion, and

n = sample size.

If 100(1 - α) % Confidence Interval for the population proportion, p, holds a particular hypothesized value of p, say p₀,

it can concluded with 100(1 - α) % confidence that the population proportion is p₀.……………. (2)

Now to work out the solution,

The given confidence interval is: 0.12 ± 0.05 i.e., [0.07, 0.17] or [7%, 17%]……………………. (3)

Part (a)

Vide (2) and (3), [7%, 17%] holds 15%. Hence the statement is NOT valid. Answer

Part (b)

Vide (2) and (3), [7%, 17%] holds 15%. Hence the statement is valid. Answer

Part (c)

Margin of error is: Z_α/2[√{p_cap(1 –p_cap)/n}]

If the sample sizes under Part (c) and (a) are respectively, n₂ and n₁, the ratio of respective margin of errors would be [Z_α/2.√t{p_cap(1 –p_cap)/n₂}]/[Z_α/2√sq.rt{p_cap(1 –p_cap)/n₁}]

= √(n₁/n₂)

= 1/√6 [given the second sample size is 6 times the first sample size]

= 0.4082

= > the margin of error under Part (c) = the margin of error under Part (c) x 0.4082

= 0.05 x 0.4082

= 0.020.

Thus, the margin of error when sample size if 6-fold is: 0.020 Answer

Part (d)

From Part (c), the new Confidence Interval is: 0.12 ± 0.02 or [0.10, 0.14], which does not hold 0.15 and hence vide (2), the conclusion would be that the program is NOT working as intended. Answer