Question

calcium is essential to tree growth. in 1990, the concentration of calcium in precipitation in chautauqua, new york, was 0.11 milligram per liter ($\frac{mg}{l}$). a random sample of 8 precipitation dates in 2018 results in the following data: 0.070, 0.108, 0.120, 0.262, 0.065, 0.234, 0.126, 0.087. a normal probability plot suggests the data could come from a population that is normally distributed. a boxplot does not show any outliers. does the sample evidence suggest that calcium concentrations have changed since 1990? use the $\alpha = 0.05$ level of significance. what are the null and alternative hypotheses? $h_0:\mu = 0.11$, $h_1:\mu \
eq 0.11$ (type integers or decimals. do not round.) find the test statistic. $t_0=\square$ (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample mean

The formula for the sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
Given data \(x = [0.070,0.108,0.120,0.262,0.065,0.234,0.126,0.087]\), \(n = 8\).
\(\sum_{i=1}^{8}x_{i}=0.070 + 0.108+0.120 + 0.262+0.065+0.234+0.126+0.087=1.072\)
\(\bar{x}=\frac{1.072}{8}=0.134\)

Step2: Calculate the sample standard deviation

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(0.070 - 0.134)^{2}=(- 0.064)^{2}=0.004096\)
\((x_{2}-\bar{x})^{2}=(0.108 - 0.134)^{2}=(-0.026)^{2}=0.000676\)
\((x_{3}-\bar{x})^{2}=(0.120 - 0.134)^{2}=(-0.014)^{2}=0.000196\)
\((x_{4}-\bar{x})^{2}=(0.262 - 0.134)^{2}=(0.128)^{2}=0.016384\)
\((x_{5}-\bar{x})^{2}=(0.065 - 0.134)^{2}=(-0.069)^{2}=0.004761\)
\((x_{6}-\bar{x})^{2}=(0.234 - 0.134)^{2}=(0.1)^{2}=0.01\)
\((x_{7}-\bar{x})^{2}=(0.126 - 0.134)^{2}=(-0.008)^{2}=0.000064\)
\((x_{8}-\bar{x})^{2}=(0.087 - 0.134)^{2}=(-0.047)^{2}=0.002209\)
\(\sum_{i = 1}^{8}(x_{i}-\bar{x})^{2}=0.004096+0.000676 + 0.000196+0.016384+0.004761+0.01+0.000064+0.002209=0.038386\)
\(s=\sqrt{\frac{0.038386}{8 - 1}}=\sqrt{\frac{0.038386}{7}}\approx0.074\)

Step3: Calculate the test - statistic

The formula for the \(t\) - test statistic is \(t_{0}=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)
Given \(\mu = 0.11\), \(\bar{x}=0.134\), \(s = 0.074\), \(n = 8\)
\(t_{0}=\frac{0.134-0.11}{0.074/\sqrt{8}}=\frac{0.024}{0.074/2.828}\)
\(t_{0}=\frac{0.024}{0.026}\approx0.92\)

Answer:

\(t_{0}=0.92\)