Question

6.128 Are Florida Lakes Acidic or Alkaline? The \(\mathrm{pH}\) of a liquid is a measure of its acidity or alkalinity. Pure water has a pH of \(7,\) which is neutral. Solutions with a pH less than 7 are acidic while solutions with a \(\mathrm{pH}\) greater than 7 are basic or alkaline. The dataset ForidaLakes givesinformation, including pH values, for a sample of lakes in Florida. Computer output of descriptive statistics for the \(\mathrm{pH}\) variable is shown:(a) How many lakes are included in the dataset? What is the mean pH value? What is the standard deviation? (b) Use the descriptive statistics above to conduct a hypothesis test to determine whether there is evidence that average pH in Florida lakes is different from the neutral value of 7 . Show all details of the test and use a \(5 \%\) significance level. If there is evidence that it is not neutral, does the mean appear to be more acidic or more alkaline? (c) Compare the test statistic and p-value you found in part (b) to the computer output below for the same data: One-Sample T: pH Test of \(m u=7\) vs not \(=7\) Variable \(\begin{array}{rrrr}\mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 53 & 6.591 & 1.288 & 0.177\end{array}\) \(\mathrm{pH}\) \(\begin{array}{crr}95 \% \mathrm{Cl} & \mathrm{T} & \mathrm{P} \\\ {[6.235,6.946)} & -2.31 & 0.025\end{array}\) 16

Short answer

From the computation, the calculated t-value suggests the mean pH is significantly different from 7. The conclusion based on the P-value and the t-value is that there is enough evidence to reject the null hypothesis at a 5% level of significance, suggesting the average pH in Florida lakes is different from the neutral value of 7. The mean pH value less than 7 indicates that the lakes are more acidic than neutral.

Step-by-step solution

Understand the given data

The given dataset represents the pH levels of some florida lakes. We know from the data that there are 53 lakes (N=53), the mean pH level is 6.591 or roughly 6.59, and the standard deviation is 1.288 or approximately 1.29.

Conduct a hypothesis test

We need to set up the null and alternative hypothesis for the test. The null hypothesis assumes the mean pH of Florida lakes, \( \mu \), is 7 (neutral). So, \( H_0 : \mu = 7 \). The alternative hypothesis assumes the mean pH is different from 7. So, \( H_a : \mu \neq 7 \). This is a two-tailed test. Since we know the mean and the standard deviation, we can calculate the test statistic (t-score) using the formula, \( t = (\bar{x} - \mu) / (\sigma / \sqrt N) \), where \( \bar{x} \) is the sample mean, \( \mu \) is the claimed population mean in the null hypothesis, \( \sigma \) is the standard deviation and N is the number of values in the sample. Inserting the given values into the formula, we can calculate \( t \). Next, we compare the calculated t-value with the table value to decide whether to reject or not to reject the null hypothesis.

Compare the test statistic and p-value

After calculating, we then compare the test statistic and p-value with the computer output. We look at the t-value and the P-value to draw up conclusions about the hypothesis.