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A chain of stores knows that 90\%90%90, percent of its customers pay with a credit or debit card, 5\%5%5, percent pay with cash, and 5\%5%5, percent pay with a service on their smartphones. They installed a new system for receiving payments, and they wonder if these percentages still hold true. They plan to take a random sample of customers in order to perform a \chi^2χ
2
\chi, squared goodness-of-fit test on the results.
What is the smallest sample size the company can take to pass the large counts condition?

Sagot :

Krishnaanil38199idn

100 people cuz 90% of 100 is 90

5 % is 5

So its easier to interpret

Answer:

The smallest sample size the company can take and still pass the large counts condition is 100 total customers.

Step-by-step explanation:

The large counts condition says that all expected counts need to be at least 5. The company needs to sample enough customers so that they expect each mode of payment to appear at least 5 times. Since the distribution they expects is uneven, we should look at the mode of payment that has the lowest expected percentage.

The lowest expected percentage is 5%, percent, which is the percentage of both cash and payment via smartphone, We need to pick a sample so 5%, percent of its size equals at least 5.

0.05x ≥5

x ≥100

So, the company needs to sample at least 100 customers.

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