Final answer:
The meteorologist's calculation of the 95% confidence interval for the mean temperature at noon over one year is correct. The formula applied uses the sample mean, the z-score for a 95% confidence level, and the sample standard deviation divided by the square root of the sample size, assuming other statistical assumptions are met.
Step-by-step explanation:
The question pertains to whether a meteorologist's 95% confidence interval for the mean temperature at noon over one year is correctly computed. The calculation involves the average temperature and the standard deviation of the temperatures recorded at noon each day for 365 days. The confidence interval is intended to estimate the true mean temperature at noon for the population of all days, not just the sample year. The formula used to calculate a 95% confidence interval is the sample mean ± (z-score)*(standard deviation)/√(sample size).
Step-by-step Calculation
The meteorologist calculates the confidence interval as 57±(1.96)*(20)/√365. This formula is derived from the Central Limit Theorem, which allows us to approximate the sampling distribution of the sample mean with a normal distribution because the sample size is large (n > 30). Since the standard deviation is given, we can assume the population standard deviation is known.
To confirm the calculation:
- The sample mean is given as 57°F.
- The z-score for a 95% confidence level is correctly used as 1.96.
- The standard deviation is 20°F.
- The sample size is 365 days.
Applying the numbers to the confidence interval formula gives us:
57 ± (1.96)*(20)/√365 = 57 ± (1.96)*(20)/19.1049 = 57 ± (1.96)*(1.0452) = 57 ± 2.0486 approximately
Therefore, the meteorologist's calculation is indeed correct provided that other assumptions required for the calculation (like the distribution of temperatures being approximately normal) are met.