Question

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 0.15 milligrams per liter​ (mg/L). A random sample of 10 precipitation dates in 2007 results in the following data table Complete parts​ (a) through​ (c) below

(a) State the hypotheses for determining if the mean concentration of calcium precipitation has changed since 1990.

​(b) Construct a 99​% confidence interval about the sample mean concentration of calcium precipitation.

​(c) Does the sample evidence suggest that calcium concentrations have changed since​ 1990?

data table.

0.073

0.081

0.052

0.273

0.125

0.179

0.137

0.245

0.319

0.095

Answer 1

(a) We state the null and alternative hypotheses for determining if the mean concentration of calcium precipitation has changed since 1990 is as below

Ho : There is not sufficient the sample evidence suggest that calcium concentrations have changed since​ 1990.

H1 : There is sufficient the sample evidence suggest that calcium concentrations have changed since​ 1990.

Using parameter the null and alternative hypotheses is

​(b) We construct a 99​% confidence interval about the sample mean concentration of calcium precipitation using following formula

We find sample mean and sample standard deviation of sample data using following table

0.073	0.1579	0.007208
0.081	0.1579	0.005914
0.052	0.1579	0.011215
0.273	0.1579	0.013248
0.125	0.1579	0.001082
0.179	0.1579	0.000445
0.137	0.1579	0.000437
0.245	0.1579	0.007586
0.319	0.1579	0.025953
0.095	0.1579	0.003956
1.579		0.077045

sample mean is

sample mean = =

sample standard deviation is

(Round answer up to 4 decimal places)

n = 10

Degrees of freedom = df = n - 1 = 10 - 1 = 9

At 99% confidence level the t is,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Using all values we get 99​% confidence interval about the sample mean concentration of calcium precipitation

We write above  99​% confidence interval using 4 decimal places is

The 99​% confidence interval about the sample mean concentration of calcium precipitation is (0.0628, 0.2530)

(c) We find that does the sample evidence suggest that calcium concentrations have changed since​ 1990.

The test statistic is

Using above all values

(Round answer up to 4 decimal places)

The test statistic = -----------(1)

Now, we find critical value of above test statistic using t table

n = 10

Degrees of freedom = df = n - 1 = 10 - 1 = 9

At 99% confidence level the t is,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

But the above test is two tailed test, then we get two critical values one with negative sign -3.250 to the left side and other with positive sign +3.250 to the right side.

critical values of t = ----------(2)

We comparing the test statistic t with it's critical values and take decision as following way

If test statistic lies outside the critical values of t then we reject the null hypothesis otherwise we fail to reject the null hypothesis.

Using equation (1) and (2)

The test statistic = lies within the critical values of t =

So, we fail to reject the null hypothesis.

There is not sufficient the sample evidence suggest that calcium concentrations have changed since​ 1990.