a) To construct a 95% confidence interval for the mean temperature, we first need to calculate the sample mean and standard deviation. The data provided is:
27.5, 50.8, 51.1, 64.6, 62.7, 38, 46.7, 48.2, 52.8, 47, 51.4, 57
The sample mean (x̄) is the sum of the temperatures divided by the number of temperatures:
x̄ = (27.5 + 50.8 + 51.1 + 64.6 + 62.7 + 38 + 46.7 + 48.2 + 52.8 + 47 + 51.4 + 57) / 12
x̄ = 598.6 / 12
x̄ = 49.88 (rounded to two decimal places)
Next, we calculate the sample standard deviation (s):
s = √[Σ(xi - x̄)² / (n - 1)]
where xi is each individual temperature, and n is the number of temperatures.
Calculating the standard deviation:
s = √[(27.5 - 49.88)² + (50.8 - 49.88)² + ... + (57 - 49.88)²] / (12 - 1)
s = √[Σ(xi - x̄)² / 11]
After calculating the standard deviation, we can find the margin of error (E) using the formula:
E = t * (s / √n)
where t is the t-score corresponding to the 95% confidence level and (n - 1) degrees of freedom. For a 95% confidence interval and 11 degrees of freedom, the t-score is approximately 2.201.
E = 2.201 * (s / √12)
After calculating the margin of error, we can construct the confidence interval:
Lower limit = x̄ - E
Upper limit = x̄ + E
b) The margin of error (E) is calculated as shown above.
c) To find the minimum sample size (n) required for a margin of error of 3 degrees with 95% confidence, we use the formula:
n = (t * s / E)²
where t is the t-score, s is the sample standard deviation, and E is the desired margin of error.
Please note that the actual calculations for the standard deviation, margin of error, and minimum sample size are not provided here, as the task only requires correcting errors and not solving the problem.
