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Assuming the population has an approximate normal distribution, if a sample size n=25

has a sample mean x=39

with a sample standard deviation s=3
, find the margin of error at a 95% confidence level. Round the answer to two decimal places.

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Answer:

Explanation:

To find the margin of error at a 95% confidence level, we need to use a t-distribution since the sample size is small (n = 25).

The formula for the margin of error with a t-distribution is:

margin of error = t*(s/sqrt(n)),

where t is the critical value from the t-distribution with n-1 degrees of freedom at a 95% confidence level (0.025 on each tail), s is the sample standard deviation, and n is the sample size.

First, we need to find the critical value from the t-distribution. Since we have n-1 = 25-1 = 24 degrees of freedom, we can find the critical value using a t-table or calculator. For a two-tailed test at a 95% confidence level, the critical value is approximately 2.064.

Next, we can plug in the values we have into the formula:

margin of error = 2.064*(3/sqrt(25))

margin of error = 1.2408

Rounding to two decimal places, the margin of error at a 95% confidence level is approximately 1.24. Therefore, we can say with 95% confidence that the true population mean is within 1.24 units of the sample mean (39).

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