The problem asks to construct a 99% confidence interval for the average land temperature from 1981-2017 and compare it to the average land temperature from 1951-1980, which was 8.79 degrees Celsius. The goal is to determine if there has been a significant change in recent years.

1. Problem Description

The problem asks to construct a 99% confidence interval for the average land temperature from 1981-2017 and compare it to the average land temperature from 1951-1980, which was 8.79 degrees Celsius. The goal is to determine if there has been a significant change in recent years.

2. Solution Steps

First, calculate the sample mean $(\bar{x})$ and sample standard deviation $(s)$ of the temperature data from 1981 to

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7. Based on the table, we have $n = 2017 - 1981 + 1 = 37$ data points.

The temperature data is: 9.301, 8.788, 9.173, 8.824, 8.799, 8.985, 9.141, 9.345, 9.076, 9.378, 9.336, 8.974, 9.008, 9.175, 9.484, 9.168, 9.326, 9.66, 9.406, 9.332, 9.542, 9.695, 9.649, 9.451, 9.829, 9.662, 9.876, 9.581, 9.657, 9.828, 9.65, 9.635, 9.753, 9.714, 9.962, 10.16, 10.

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We calculate the sample mean:

$\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$

$\bar{x} = \frac{9.301 + 8.788 + ... + 10.049}{37} = \frac{345.704}{37} \approx 9.343$

Next, we calculate the sample standard deviation:

$s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}}$

First compute $\sum_{i=1}^{n} (x_i - \bar{x})^2$

$\sum_{i=1}^{n} (x_i - \bar{x})^2 = (9.301-9.343)^2 + (8.788-9.343)^2 + ... + (10.049-9.343)^2 = 2.156$

$s = \sqrt{\frac{2.156}{37-1}} = \sqrt{\frac{2.156}{36}} = \sqrt{0.0599} \approx 0.245$

Now, construct the 99% confidence interval for the population mean $\mu$.

Since the sample size is relatively small ($n = 37 < 40$), we will use the t-distribution.

The confidence level is 99%, so $\alpha = 1 - 0.99 = 0.01$. Since it is a two-tailed test, $\alpha/2 = 0.005$. The degrees of freedom are $df = n - 1 = 37 - 1 = 36$.

The critical t-value, $t_{\alpha/2, df} = t_{0.005, 36} \approx 2.719$ (from a t-table or calculator).

The margin of error is:

$E = t_{\alpha/2, df} \cdot \frac{s}{\sqrt{n}} = 2.719 \cdot \frac{0.245}{\sqrt{37}} = 2.719 \cdot \frac{0.245}{6.083} \approx 2.719 \cdot 0.0403 \approx 0.1096$

The 99% confidence interval is:

$(\bar{x} - E, \bar{x} + E) = (9.343 - 0.1096, 9.343 + 0.1096) = (9.2334, 9.4526)$

Since 8.79 is not within the interval (9.2334, 9.4526), there is a significant difference between the average land temperature from 1981-2017 and the average land temperature from 1951-

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3. Final Answer

The 99% confidence interval for the average land temperature from 1981-2017 is (9.2334, 9.4526) degrees Celsius. Since 8.79 degrees Celsius (the average from 1951-1980) is not within this interval, there is a significant difference.