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A store manager claims that 60% of shoppers who enter her store make a purchase. To investigate this claim, she selects a random sample of 40 customers and finds that 40% of them make a purchase. She wants to know if the data provide convincing evidence that the true proportion of all customers who enter her store that will make a purchase differs from 60%. The P-value of this test is 0.0098. Interpret the P-value.

Assuming the true population proportion of shoppers who will make a purchase is 0.60, the probability of obtaining a sample statistic of 0.40 in a random sample of 40 customers is 0.0098.
Assuming the true population proportion of shoppers who will make a purchase differs from 0.60, the probability of obtaining a sample statistic of 0.40 in a random sample of 40 customers is 0.0098.
Assuming the true population proportion of shoppers who will make a purchase is 0.60, the probability of obtaining a sample statistic as extreme as or more extreme than 0.40 in a random sample of 40 customers is 0.0098.
Assuming the true population proportion of shoppers who will make a purchase is different from 0.60, the probability of obtaining a sample statistic as extreme as or more extreme than 0.40 in a random sample of 40 customers is 0.0098.

Sagot :

Answer:

I do believe that it is Option B

(Assuming the true population proportion of shoppers who will make a purchase differs from 0.60, the probability of obtaining a sample statistic of 0.40 in a random sample of 40 customers is 0.0098.)

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