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Harman [31]
3 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]3 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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