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SIZIF [17.4K]
3 years ago
6

Suppose the probability of an IRS audit is 1.5 percent for U.S. taxpayers who file form 1040 and who earned $100,000 or more.

Mathematics
1 answer:
sp2606 [1]3 years ago
8 0

Answer:

(A) The odds that the taxpayer will be audited is approximately 0.015.

(B) The odds against these taxpayer being audited is approximately 65.67.

Step-by-step explanation:

The complete question is:

Suppose the probability of an IRS audit is 1.5 percent for U.S. taxpayers who file form 1040 and who earned $100,000 or more.

A. What are the odds that the taxpayer will be audited?

B. What are the odds against such tax payer being audited?

Solution:

The proportion of U.S. taxpayers who were audited is:

P (A) = 0.015

Then the proportion of U.S. taxpayers who were not audited will be:

P (A') = 1 - P (A)

        = 1 - 0.015

        = 0.985

(A)

Compute the  odds that the taxpayer will be audited as follows:

\text{Odds of being Audited}=\frac{P(A)}{P(A')}

                                    =\frac{0.015}{0.985}\\\\=\frac{3}{197}\\\\=0.015228\\\\\approx 0.015

Thus, the odds that the taxpayer will be audited is approximately 0.015.

(B)

Compute the odds against these taxpayer being audited as follows:

\text{Odds against Audited}=\frac{P(A')}{P(A)}

                                    =\frac{0.985}{0.015}\\\\=\frac{3}{197}\\\\=65.666667\\\\\approx 65.67

Thus, the odds against these taxpayer being audited is approximately 65.67.

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Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>

SO, Null Hypothesis, H_0 : p = 25%   {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of 600 cars in San Francisco who are blue =   \frac{141}{600} = 0.235

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So, <u><em>test statistics</em></u>  =  \frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }

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<em>Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

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