States is 2.24. Assume the population standard deviation number of television sets in the United St is 1.38 a) Describe the sampling distribution for all sample sizes of n = 40. The distribution will be normal because the sample size is greater than 30 b) A random sample of 40 households results in a sample mean of 2.55 tv sets per household. Wh. the probability of obtaining a sample mean of at least 2.55 if the population mean is 2.24? Does contradict the results reported by A.C. Nielsen? 2.55-2.24 - 0.046 Since the probability is larger than 0.05 it is not unusual 1.38 V40 1-P(Z>0.05) = 0.52 c) construct a 95% confidence interval for the sample mean. How does this support your conclus on part b? (2.55+1.96() 1.38 V40- (2.55-0.428),(2.55+0.428) (2.12, 2.98) It supports b because mean 2.55 lies between 2.12and 2.98 d) Which would be more likely to happen...one household to have 3 televisions or a sample o households to have a mean of 3 televisions? Explain.

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I need subpart (d) please 

1) Based on data obtained from A.C. Nielsen, the mean # of televisions in a household in the United
States is 2.24. Assume the population standard deviation number of television sets in the United States
is 1.38
a) Describe the sampling distribution for all sample sizes of n = 40.
The distribution will be normal because the sample size is greater than 30
b) A random sample of 40 households results in a sample mean of 2.55 tv sets per household. What is
the probability of obtaining a sample mean of at least 2.55 if the population mean is 2.24? Does this
contradict the results reported by A.C. Nielsen?
2.55-2.24
- 0.046
Since the probability is larger than 0.05 it is not unusual
1.38
V40
1-P(Z>0.05)
= 0.52
c) construct a 95% confidence interval for the sample mean. How does this support your conclusion
on part b?
(2.55+1.96()
1.38
V40
(2.55-0.428),(2.55+0.428)
(2.12, 2.98)
It supports b because mean 2.55 lies between 2.12and 2.98
d) Which would be more likely to happen...one household to have 3 televisions or a sample of 40
households to have a mean of 3 televisions? Explain.
Transcribed Image Text:1) Based on data obtained from A.C. Nielsen, the mean # of televisions in a household in the United States is 2.24. Assume the population standard deviation number of television sets in the United States is 1.38 a) Describe the sampling distribution for all sample sizes of n = 40. The distribution will be normal because the sample size is greater than 30 b) A random sample of 40 households results in a sample mean of 2.55 tv sets per household. What is the probability of obtaining a sample mean of at least 2.55 if the population mean is 2.24? Does this contradict the results reported by A.C. Nielsen? 2.55-2.24 - 0.046 Since the probability is larger than 0.05 it is not unusual 1.38 V40 1-P(Z>0.05) = 0.52 c) construct a 95% confidence interval for the sample mean. How does this support your conclusion on part b? (2.55+1.96() 1.38 V40 (2.55-0.428),(2.55+0.428) (2.12, 2.98) It supports b because mean 2.55 lies between 2.12and 2.98 d) Which would be more likely to happen...one household to have 3 televisions or a sample of 40 households to have a mean of 3 televisions? Explain.
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