Calcium in RainwaterCalcium is essential to tree growth. In 1990, the concentration of calcium in precipitation in Chautauqua, New York, was 0.11 milligram per liter (mg/L). A random sample of 10 precipitation dates in 2018 results in the following data:



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 α = 0.05 level of significance.

Verified step by step guidance

1. Step 1: Define the hypotheses for the test. The null hypothesis \(H_0\) is that the mean calcium concentration in 2018 is equal to the 1990 concentration, i.e., \(\mu = 0.11\) mg/L. The alternative hypothesis \(H_a\) is that the mean calcium concentration has changed, i.e., \(\mu \neq 0.11\) mg/L.
2. Step 2: Calculate the sample mean \(\bar{x}\) and the sample standard deviation \(s\) from the given data values: 0.065, 0.087, 0.070, 0.262, 0.126, 0.183, 0.120, 0.234, 0.313, and 0.108 mg/L.
3. Step 3: Determine the test statistic using the one-sample t-test formula: \(\displaystyle t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}\) where \(\mu_0 = 0.11\) mg/L, \(n=10\) is the sample size, \(\bar{x}\) is the sample mean, and \(s\) is the sample standard deviation.
4. Step 4: Find the critical t-value(s) for a two-tailed test at the \(\alpha = 0.05\) significance level with \(n-1 = 9\) degrees of freedom using a t-distribution table or calculator.
5. Step 5: Compare the calculated t-statistic to the critical t-value(s). If the absolute value of the t-statistic is greater than the critical value, reject the null hypothesis, indicating evidence that calcium concentrations have changed since 1990. Otherwise, do not reject the null hypothesis.

Key Concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to decide whether there is enough evidence to reject a null hypothesis. In this context, the null hypothesis states that the calcium concentration in 2018 is equal to that in 1990 (0.11 mg/L). The alternative hypothesis suggests a change in concentration. The test uses sample data and a significance level (α = 0.05) to make this decision.

One-Sample t-Test

A one-sample t-test compares the sample mean to a known population mean when the population standard deviation is unknown. It is appropriate here because the sample size is small (n=10) and the data is approximately normally distributed. The test calculates a t-statistic to determine if the sample mean significantly differs from 0.11 mg/L.

Normality Assumption and Outliers

The validity of the t-test depends on the assumption that the data come from a normally distributed population. The normal probability plot and boxplot help assess this assumption. Since the plot suggests normality and no outliers are present, the t-test results are reliable for this data set.