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Sagot :
Answer:
a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.
b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.
c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.
d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.
This means that [tex]\mu = 273, \sigma = 100[/tex]
A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?
Sample of 30 means that [tex]n = 30, s = \frac{100}{\sqrt{30}}[/tex]
The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So
X = 289
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}[/tex]
[tex]Z = 0.88[/tex]
[tex]Z = 0.88[/tex] has a p-value of 0.8106
X = 257
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}[/tex]
[tex]Z = -0.88[/tex]
[tex]Z = -0.88[/tex] has a p-value of 0.1894
0.8106 - 0.1894 = 0.6212
0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.
B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?
Sample of 30 means that [tex]n = 50, s = \frac{100}{\sqrt{50}}[/tex]
X = 289
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}[/tex]
[tex]Z = 1.13[/tex]
[tex]Z = 1.13[/tex] has a p-value of 0.8708
X = 257
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}[/tex]
[tex]Z = -1.13[/tex]
[tex]Z = -1.13[/tex] has a p-value of 0.1292
0.8708 - 0.1292 = 0.7416
0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.
C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?
Sample of 30 means that [tex]n = 100, s = \frac{100}{\sqrt{100}}[/tex]
X = 289
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}[/tex]
[tex]Z = 1.6[/tex]
[tex]Z = 1.6[/tex] has a p-value of 0.9452
X = 257
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}[/tex]
[tex]Z = -1.6[/tex]
[tex]Z = -1.6[/tex] has a p-value of 0.0648
0.9452 - 0.0648 =
0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.
D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?
None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.
C is actually 0.8904
for anybody else stuck on this wondering why cengage is telling you c is wrong
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