You are the assistant manager at Extra Credit Union. Your boss, Robin Banks, has sent you the following email: "Assistant Manager, as you know, we go the extra mile in our service to our clients. Part of that is knowing our clients well. Thus, we are wanting to solicit phone-call feedback from some of our clients. I know we don't have time to talk to all 10,000 clients, so I am wanting a smaller representative sample. Perhaps a random sample of 25 or 50 would be good. Across all 10,000 clients, the average account balance is $12,000 and the standard deviation is $5000. I want the sample we take to closely match our clients as a whole. Specifically, I want the probability to be at least 80% that the average of our sample is within $1000 of the average for all clients. Would a random sample

Winston Todd

Winston Todd

Answered question

2022-10-16

You are the assistant manager at Extra Credit Union. Your boss, Robin Banks, has sent you the following email:
"Assistant Manager, as you know, we go the extra mile in our service to our clients. Part of that is knowing our clients well. Thus, we are wanting to solicit phone-call feedback from some of our clients. I know we don't have time to talk to all 10,000 clients, so I am wanting a smaller representative sample. Perhaps a random sample of 25 or 50 would be good. Across all 10,000 clients, the average account balance is $12,000 and the standard deviation is $5000. I want the sample we take to closely match our clients as a whole. Specifically, I want the probability to be at least 80% that the average of our sample is within $1000 of the average for all clients. Would a random sample of 25 be fine or do we need to take a random sample of 50? Use the magical statistics skills you learned back in college and get back to me ASAP! Thanks, Robin Banks"
In your response indicate whether 25 is sufficient or whether a sample of 50 would be necessary. (You will want to calculate the probability of the sample mean being between $11,000 and $13,000 for both a sample size of 25 and 50.)

Answer & Explanation

Cavalascamq

Cavalascamq

Beginner2022-10-17Added 21 answers

Answer:
Confidence level =0.8 and population S D = 5000 and ϵ = 1000 z critical corresponding to 0.8 is 1.28
n = ( Z C S D / E ) 2 n = ( 1.28 5000 / 1000 ) 2 n = 40.96 n = 41
Here from calculation we got n=41 for the probability to be at least 80% that the average of Sample is within $1000 of the average for all clients

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