# Calcium is essential to tree growth because it promotes the formation of wood and maintains cell walls. In 1990, the concentration of calcium in precipitation in a certain area was 0.11 milligrams per liter (mg/L). A random sample of 10 precipitation dates in 2007 results in the following data table:

0.077, 0.071, 0.081, 0.257, 0.116, 0.185, 0.103, 0.223, 0.318, 0.116

a) Construct a 90% confidence interval about the sample mean concentration of calcium precipitation:

(____ , ____) (use ascending order, round to four decimal places)

b) Does the sample evidence suggest that calcium concentrations have changed since 1990?

(Choose one from below)

1) Yes, because the confidence intervalcontains 0.11.

2) No, beacuse the confidence interval does not contain 0.11.

3) Yes, beacuse the confidence interval does not contain 0.11.

4) No, because the confidence interval contains 0.11.

## a) (0.1020, 0.1445)

b) 3) Yes, beacuse the confidence interval does not contain 0.11.

## To construct a confidence interval, we need to calculate the sample mean and the margin of error.

a) First, let's calculate the sample mean concentration. Add up all the values in the data table and divide by the number of observations (n=10):

0.077 + 0.071 + 0.081 + 0.257 + 0.116 + 0.185 + 0.103 + 0.223 + 0.318 + 0.116 = 1.547

Sample mean concentration = 1.547 / 10 = 0.1547 (rounded to four decimal places)

Next, we need to calculate the margin of error. To do this, we use the standard error formula:

Standard Error = standard deviation / square root of sample size

Since we only have a sample, we need to estimate the population standard deviation from the sample standard deviation. The formula for the sample standard deviation is:

Sample Standard Deviation = square root of [Σ(xi - x̄)² / (n - 1)]

Using the values from the data table, we calculate:

(0.077 - 0.1547)² + (0.071 - 0.1547)² + (0.081 - 0.1547)² + (0.257 - 0.1547)² + (0.116 - 0.1547)² + (0.185 - 0.1547)² + (0.103 - 0.1547)² + (0.223 - 0.1547)² + (0.318 - 0.1547)² + (0.116 - 0.1547)² = 0.191947

Sample Standard Deviation = square root of (0.191947 / (10 - 1)) = 0.154468 (rounded to six decimal places)

Now we can calculate the standard error:

Standard Error = 0.154468 / square root of 10 = 0.048779 (rounded to six decimal places)

To find the margin of error, multiply the standard error by the appropriate critical value from the t-distribution. Since we want a 90% confidence interval, we look up the critical value for a two-tailed test with 10 - 1 = 9 degrees of freedom. From the t-distribution table, the critical value is approximately 1.833.

Margin of Error = 1.833 x 0.048779 = 0.089361 (rounded to six decimal places)

Finally, construct the confidence interval by subtracting and adding the margin of error to the sample mean:

Confidence Interval = (Sample mean concentration - Margin of Error, Sample mean concentration + Margin of Error)

Confidence Interval = (0.1547 - 0.089361, 0.1547 + 0.089361)

Confidence Interval = (0.065339, 0.244061) (rounded to four decimal places)

Therefore, the 90% confidence interval about the sample mean concentration of calcium precipitation is (0.0653, 0.2441).

b) Now let's determine if the sample evidence suggests that calcium concentrations have changed since 1990. We compare the confidence interval to the 1990 concentration of 0.11 mg/L.

Since the confidence interval (0.0653, 0.2441) does not contain 0.11, the correct answer is:

3) Yes, because the confidence interval does not contain 0.11.

This means that based on the sample evidence, there is reason to believe that calcium concentrations have changed since 1990.

## To construct a 90% confidence interval for the sample mean concentration of calcium precipitation in 2007, we can use the formula:

Confidence interval = sample mean ± (t-value * standard deviation / √n)

Step 1: Calculate the sample mean

The sample mean (x̄) can be calculated by adding up all the observations and dividing by the sample size (n):

x̄ = (0.077 + 0.071 + 0.081 + 0.257 + 0.116 + 0.185 + 0.103 + 0.223 + 0.318 + 0.116) / 10

x̄ = 1.547 / 10

x̄ = 0.1547 (rounded to four decimal places)

Step 2: Calculate the standard deviation (s)

First, calculate the sum of the squared differences between each observation and the sample mean:

(0.077 - 0.1547)^2 + (0.071 - 0.1547)^2 + (0.081 - 0.1547)^2 + (0.257 - 0.1547)^2 + (0.116 - 0.1547)^2 + (0.185 - 0.1547)^2 + (0.103 - 0.1547)^2 + (0.223 - 0.1547)^2 + (0.318 - 0.1547)^2 + (0.116 - 0.1547)^2

= 0.006721 + 0.007201 + 0.007661 + 0.009526 + 0.001504 + 0.000926 + 0.007348 + 0.005520 + 0.017801 + 0.001504

= 0.066130

Next, divide the sum by (n-1) to get the sample variance:

s^2 = 0.066130 / (10-1)

s^2 = 0.007348

Finally, take the square root to get the standard deviation:

s = √0.007348

s ≈ 0.0858 (rounded to four decimal places)

Step 3: Find the t-value

To find the t-value, we need the degrees of freedom (df), which is equal to (n-1):

df = 10-1

df = 9

Using a t-table or calculator, for a 90% confidence level and 9 degrees of freedom, the t-value is approximately 1.833.

Step 4: Calculate the confidence interval

Substituting the values into the formula:

Confidence interval = 0.1547 ± (1.833 * 0.0858 / √10)

Calculating the components:

1.833 * 0.0858 ≈ 0.157 (rounded to three decimal places)

√10 ≈ 3.162 (rounded to three decimal places)

Finally:

Confidence interval = 0.1547 ± (0.157 / 3.162)

Confidence interval ≈ (0.1547 - 0.0496, 0.1547 + 0.0496)

Confidence interval ≈ (0.1051, 0.2043) (rounded to four decimal places)

(a) The 90% confidence interval about the sample mean concentration of calcium precipitation is approximately (0.1051, 0.2043) (in ascending order).

(b) The correct answer is 2) No, because the confidence interval does not contain 0.11. The confidence interval obtained does not contain the value 0.11, which was the concentration in 1990. This suggests that there may have been a change in calcium concentrations since 1990.