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customers of this company, describe the sampling distribution of the sample proportion of churn.SD(p)=√PQn¿√(0.25) (1−0.25)200¿0.03061862178479b.What is the probability that a simple random sample of 200 customers will have a sample proportion of churn that is within ± 0.04 of the population proportion?P(0.21<p<0.29)¿P(0.21−0.250.03061862178479)<z<(0.29−0.250.03061862178479)
Unit 6 Exercises3¿P(−1.306394529<z<0.765465545)=0.7794−0.0951¿0.6843c.Check the condition for using the normal distribution to approximate the sampling distribution of the sample proportion and see whether it is valid.??4.A simple random sample of 100 items is taken from an inventory and the average unit cost on the items is $25.5. The population standard deviation is known to be $10.25. (σ8 points)a.Construct a 95% confidence interval for the mean unit cost of the populationMean±z0.05/2(σ√n)¿25.5−(1.96∗(10.25√100))¿23.491¿25.5+(1.96∗(10.25√100))¿27.509¿(23.491,27.509)b.What is the margin of error at the 90% confidence level?Marginof Error=z0.1/2(σ√n)¿(1.645∗(10.25√100))¿1.686125c.Construct a 95% confidence interval for the mean unit cost of the population if the sample unit cost is obtained from a random sample of 150 instead of 100.Mean±z0.05/2(σ√n)¿25.5−(1.96∗(10.25√150))¿24.1232847
Unit 6 Exercises4¿25.5+(1.96∗(10.25√150))¿27.1403416324.1232847,27.1403416¿¿ ) d.Based on the results from a. and c., what can you say about the effect of a larger sample size on the length of the confidence interval at the same confidence level?The variance will decrease 5.A simple random sample of 60 items is taken from an inventory and the average unit cost on the items is $23.5. The population standard deviation is unknown. Instead the sample σstandard deviation s is also calculated from the sample and is found to be $12.2. (4 points)a.Construct a 99% confidence interval for the mean unit cost of the population.