A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normally distributed. Find the 95% confidence interval for the true mean. Round to the nearest cent
$3.60 $450 $2.80 $6.30 $260 $5.20 $6.75 $4.25 $8.00 $3.00
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First, we need to calculate the mean and standard deviation of the sample.
Mean (x̄) = (3.60 + 450 + 2.80 + 6.30 + 260 + 5.20 + 6.75 + 4.25 + 8.00 + 3.00) / 10 = $75.01
Next, we calculate the standard deviation.
Standard deviation (s) = sqrt[((3.60 - 75.01)^2 + (450 - 75.01)^2 + (2.80 - 75.01)^2 + (6.30 - 75.01)^2 + (260 - 75.01)^2 + (5.20 - 75.01)^2 + (6.75 - 75.01)^2 + (4.25 - 75.01)^2 + (8.00 - 75.01)^2 + (3.00 - 75.01)^2) / (10 - 1)] = $141.64
The 95% confidence interval for the true mean is given by the formula:
x̄ ± (t * s / sqrt(n))
where t is the t-value for a 95% confidence interval with n - 1 degrees of freedom. For a sample size of 10, the degrees of freedom is 10 - 1 = 9. The t-value for a 95% confidence interval with 9 degrees of freedom is approximately 2.262.
So, the 95% confidence interval for the true mean is:
75.01 ± (2.262 * 141.64 / sqrt(10)) = 75.01 ± 100.64
Therefore, the 95% confidence interval for the true mean is approximately $-25.63 to $175.65.
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