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Misha Larkins [42]
4 years ago
12

Nicholas earned 8% in his savings account. if he is in the 28% tax bracket, what is his after-tax savings rate of return?

Mathematics
1 answer:
Kamila [148]4 years ago
5 0
If Nicholas is going to pay 28% in taxes of the 8% interest he is earning, then his net percentage would be 72% of the 8%.

To determine this overall rate of return multiply 0.72 x 0.08.

0.72 x 0.08 = 0.0576 or 5.76 % rate of return.
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Answer:

The correct 95% confidence interval is (8.4, 8.8).

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The information provided is:

n=500\\\bar x=8.6\\\sigma=2.2

(a)

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

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The 95% confidence interval for the average satisfaction score is computed as:

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This confidence interval is incorrect.

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(b)

The (1 - <em>α</em>)% confidence interval for the parameter implies that there is (1 - <em>α</em>)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - <em>α</em>)% confidence interval for population parameter implies that there is a (1 - <em>α</em>) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

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According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

<em>n </em>= 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

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