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barxatty [35]
3 years ago
11

According to a survey of 500, the mean income before taxes of consumer units (i.e., households) inthe U.S. was $60,533 with a st

andard error of 717.51. Find the margin of error for constructing a95% confidence interval on the population mean income before taxes of all consumer units in theU.S
Mathematics
1 answer:
ioda3 years ago
3 0

Answer:

The margin of error for constructing a 95% confidence interval on the population mean income before taxes of all consumer units in the U.S is 1406.32.

Step-by-step explanation:

We are given that according to a survey of 500, the mean income before taxes of consumer units (i.e., households) in the U.S. was $60,533 with a standard error of 717.51.

Margin of error tells us that how much our sample mean value deviates from the true population value.

<u></u>

<u>Margin of error is calculated using the following formula;</u>

        Margin of error =  Z_(_\frac{\alpha}{2}_) \times \text{Standard of Error}

where, \alpha = level of significance = 1 - confidence level

                                                    = 1 - 0.95 = 0.05 or 5%

Standard of Error =  \frac{\sigma}{\sqrt{n} } = 717.51

Now, the value of z at 2.5% level of significance (\frac{0.05}{2} =0.025) is given in the z table as 1.96, that means;

            Margin of error =  Z_(_\frac{\alpha}{2}_) \times \text{Standard of Error}

                                      =  1.96 \times 717.51 = 1406.32

Hence, the margin of error for constructing a 95% confidence interval on the population mean income before taxes of all consumer units in the U.S is 1406.32.

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All you have to do is plug in the number for x, and find y. I will write this out as if in that chart you have shown.

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4 0
3 years ago
Read 2 more answers
A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the
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Answer:

The required sample size for the new study is 801.

Step-by-step explanation:

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