Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A researcher studies water clarity at the same location in a lake on the same da

ID: 3172130 • Letter: A

Question

A researcher studies water clarity at the same location in a lake on the same dates during the course of a year and repeats the measurements on the same dates 5 years later. The researcher immerses a weighted disk painted black and white and measures the depth (in inches) at which it is no longer visible. The collected data is given in the table below. Complete parts (a) and (b) below. Why is it important to take the measurements on the same date? Those are the same dates that all biologists use to take water clarity samples. Using the same dates maximizes the difference in water clarity. Using the same dates makes it easier to remember to take samples. Using the same dates makes the second sample dependent on the first. Does the evidence suggest that the clarity of the lake is improving at the alpha = 0.05 level of significance? Note that the normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

Explanation / Answer

Part a

Answer:

D. Using the same dates makes the second sample dependent on the first.

(Second sample observations should be based first sample observations for the paired t test for the population mean.)

Part b

Here, we have to use paired t test.

H0: µ1=µ2 versus Ha: µ1 µ2

Test statistic = Dbar/[Sd/sqrt(n)]

Where, n is sample size.

Calculation table for this test is given as below:

Before

After

Di

(Di - DBar)^2

65.2

62.6

2.6

30.06694444

61.7

60.6

1.1

15.86694444

42.8

51.4

-8.6

32.68027778

63.5

68.7

-5.2

5.366944444

70.2

76.6

-6.4

12.36694444

67.4

68.2

-0.8

4.340277778

DBar

-2.88333

We are given

Sample Size

6

DBar

-2.8833

Degrees of Freedom

5

SD

4.4875

Standard error = Sd/sqrt(n)

Standard error = 4.4875/sqrt(6) = 1.8320

Test statistic = -2.8833/1.8320

Test statistic = -1.573853712

P-value = 0.1763

Alpha value = 0.05

P-value > Alpha value

So, we do not reject the null hypothesis H0

Before

After

Di

(Di - DBar)^2

65.2

62.6

2.6

30.06694444

61.7

60.6

1.1

15.86694444

42.8

51.4

-8.6

32.68027778

63.5

68.7

-5.2

5.366944444

70.2

76.6

-6.4

12.36694444

67.4

68.2

-0.8

4.340277778

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote