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Harman [31]
2 years ago
13

The World Health Organization is planning a study of the average weight gain of Europeans in the last 5 years. Scientists from t

he WHO plan on taking a SRS of 100 Europeans and calculating a 95% confidence interval of the average weight gain μ of all Europeans. The WHO changed their minds and now plans on calculating a 90% confidence interval from the sample they will select. Explain to the WHO the impact of changing their confidence level from 95% to 90%.
Their confidence interval would be less likely to capture the sample mean.

They would decrease the margin of error of their confidence interval if they calculated a 90% rather than a 95% confidence interval.

The probability of selecting a sample which doesn't capture the true value of μ would be 10% rather than 5% if they decide to calculate a 90% confidence interval rather than a 95% confidence interval from the sample they will select
Mathematics
1 answer:
Vera_Pavlovna [14]2 years ago
6 0

Answer:

The probability of selecting a sample which doesn't capture the true value of μ would be 10% rather than 5% if they decide to calculate a 90% confidence interval rather than a 95% confidence interval from the sample they will select.

Step-by-step explanation:

The correct answer is the last statement, but first, let's look at the other statements:

<em>Their confidence interval would be less likely to capture the </em><em>sample </em><em>mean.</em>

This statement is not correct because the <em>confidence </em>of a confidence interval gives us the probability of capturing the true value of the population mean.

<em>They would </em><em>decrease </em><em>the margin of error of their confidence interval if they calculated a 90% rather than a 95% confidence interval.</em>

If we want more confidence we must establish more precision, which means more error. In other words if the confidence increases so does the error for a fixed sample size. The second statement is false.

<em>The probability of selecting a sample which doesn't capture the true value of μ would be 10% rather than 5% if they decide to calculate a 90% confidence interval rather than a 95% confidence interval from the sample they will select.</em>

As stated before the <em>confidence </em>of a confidence interval is the probability that the interval contains the true value of μ. Therefore, if we increase the confidence from 95% to 90% then this probability also increases. A 90% confidence means that there's a 10% probability of not containing the true mean. Likely, a 95% confidence means that there's a 5% probability of not containig the true mean. The third statement is true.

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Carlos has at most $40 to spend on food for a barbeque. He wants to buy hot dogs and hamburgers and then
poizon [28]

Answer:

$8.50

Step-by-step explanation:

I don't know what Alls are, but if Carlos insists, we can calculate how much he can spend on Alls with the expression:

($31.50) + r \leq $40

This says the sum of what Carlos has already spent (hot dogs and hamburgers) plus the amount he spends on Alls (rolls?), r,  must be equal to or less than the $40 he has allowed himself to spend.

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Mary bought these art supplies: 4 paint brushes @ $1.29 each, 6 dowel rods @ $1.49 each, a jar of paste @ $1.25, and 2 canvases
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Answer:

Option (1) is correct

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Step-by-step explanation:

Given: Mary bought the art supplies.

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Since, she brought 4 paint brushes @ $1.29 each,

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Thus, cost of 2 canvases  is $12.00

Thus, total cost of art supplies = 5.16 + 8.94 + 1.25 + 12 =$27.35

Thus, amount she receives back from $40 is $40 - $12.35 = $12.65

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