Question

The pH of 20 randomly selected lakes is measured. Their average pH is 5.7.

Part A. Historically the standard deviation in the pH values is 0.9. Use this standard deviation for the following questions.
Part Ai. Build a 95% confidence interval for the population mean lake pH.
Part Aii. Build a 90% confidence interval for the population mean lake pH.
Part Aiii. Build an 80% confidence interval for the population mean lake pH.
Part Aiv. Compare the intervals you created in Ai, Aii and Aiii. What effect does changing the level of confidence have on the interval?

Please solve only part B

Part B. For the 20 measured lakes the standard deviation in the pH values is 0.9. Use this standard deviation for the following questions.
Part Bi. Build a 95% confidence interval for the population mean lake pH.
Part Bii. Compare the confidence intervals in Bi and Ai. What effect does not knowing the value of ? have on the interval?
Part Biii. Test whether the population mean pH differs from 6.

Answer 1

Part Bi.

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 5.7

S = 0.9

n = 20

df = n – 1 = 19

Confidence level = 95%

Critical t value = 2.0930

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 5.7 ± 2.0930*0.9/sqrt(20)

Confidence interval = 5.7 ± 0.4212

Lower limit = 5.7 - 0.4212 = 5.28

Upper limit = 5.7 + 0.4212 = 6.12

Confidence interval = (5.28, 6.12)

Part Bii.

We have

Confidence interval with unknown σ = (5.28, 6.12)

Confidence interval with known σ = (5.31, 6.09)

This means, width of the confidence interval with unknown σ is more than that for known σ.

If we don’t know the population standard deviation σ, then width of the confidence interval increases.

Part Biii.

One sample t-test

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H₀: the population mean pH do not differs from 6.

Alternative hypothesis: H_a: the population mean pH differs from 6.

H₀: µ = 6 versus H_a: µ ≠ 6

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 6

Xbar = 5.7

S = 0.9

n = 20

df = n – 1 = 19

α = 0.05

Critical value = - 2.0930 and 2.0930

(by using t-table or excel)

t = (5.7 – 6)/[0.9/sqrt(20)]

t = -1.4907

P-value = 0.1525

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that the population mean pH differs from 6.