A researcher is interested in examining if the average temperature in a particular state in the US was increased due to significant changes to its landscape over the last several decades. The average yearly temperature was recorded from 50 randomly sampled locations from this state in 1948, and again in 2018. The 95% confidence interval for
𝜇
𝑑
was calculated to be (0.87,4.93). What is the sample standard error of the temperature differences? Assume all conditions for statistical inferences are satisfied.

Question

A researcher is interested in examining if the average temperature in a particular state in the US was increased due to significant changes to its landscape over the last several decades. The average yearly temperature was recorded from 50 randomly sampled locations from this state in 1948, and again in 2018. The 95% confidence interval for 
𝜇
𝑑
 was calculated to be (0.87,4.93). What is the sample standard error of the temperature differences? Assume all conditions for statistical inferences are satisfied.

BlackTom AI · Accepted Answer

The question provides a 95% confidence interval for the mean difference in temperature between 1948 and 2018: (0.87, 4.93).
First, compute the half-width (margin of error) of this interval: (4.93 − 0.87) / 2 = 2.03.
The center of the interval is (4.93 + 0.87) / 2 = 2.90, but what we need for the standard error is the margin of error.
For a 95% CI with n = 50 observations, the standard method uses t* with df = n − 1 = 49. The approximate t* value is about 2.01 (specifically around 2.0096).
Thus, MOE = t* × SE, so SE ≈ MOE / t* ≈ 2.03 / 2.01 ≈ 1.01.
The quantity SE here refers to the standard error of the mean difference, which for paired or matched differences is s_d / √n, where s_d is the sample standard deviation of the differences.
Therefore, the numerical value we obtain for the standard error of the temperature differences is approximately 1.0, which lies between 1 and 3.
Now evaluate each answer choice:
- Less than 1: This would require a margin of error smaller than about 2.0 with a t* around 2, which is inconsistent with the given interval width (the computed MOE is about 2.03). Hence this option is not supported by the data.
- Between 1 and 3: This matches the calculated SE ≈ 1.01, which falls squarely in this range.
- Between 3 and 6: The SE would need to be notably larger (roughly >3) to yield the observed MOE with t* ≈ 2, which is contradicted by the actual interval.
- Greater than 6: This would imply an extremely large SE and a very wide interval, which is clearly not the case here.
In summary, the width of the confidence interval, when translated back to a standard error using the appropriate t* value and sample size, indicates a standard error around 1.0, supporting the option in the range between 1 and 3.

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