Question

THIS PROBLEM NEEDS TO BE SOLVED ONLY USING EXCEL SOFTWARE!

1. Use the following data set to answer the following:

ew	dbh
e	23.5
e	43.5
e	6.6
e	11.5
e	17.2
e	38.7
e	2.3
e	31.5
e	10.5
e	23.7
e	13.8
e	5.2
e	31.5
e	22.1
e	6.7
e	2.6
e	6.3
e	51.1
e	5.4
e	9
e	43
e	8.7
e	22.8
e	2.9
e	22.3
e	43.8
e	48.1
e	46.5
e	39.8
e	10.9
w	17.2
w	44.6
w	44.1
w	35.5
w	51
w	21.6
w	44.1
w	11.2
w	36
w	42.1
w	3.2
w	25.5
w	36.5
w	39
w	25.9
w	20.8
w	3.2
w	57.7
w	43.3
w	58
w	21.7
w	35.6
w	30.9
w	40.6
w	30.7
w	35.6
w	18.2
w	2.9
w	20.4
w	11.4

- Make a back to back stemplot (use https://www.learner.org/courses/againstallodds/interactives/stemplots.html click "go" under "plot your data mode" and then click "back to back datasets".

- Make side by side boxplots (use http://www.imathas.com/stattools/boxplot.html)

A) Is it appropoiate to use t-procedures to compare the mean DBH of trees in the north half of the tract with the mean DBH of the trees in the south half. Are conditions met? Check them.

B)Null hypothesis? alternative hypothesis? for comparing the two samples of tree DBHs? give reasons for your choices.

C) Perform a significance tesr. Report the test statistic, the dregress of freedom, and the p-value.

D) Find 95% confidence interval for the difference mean DBHs. Explain how this interval provides additional information about this problem.

Answer 1

Data and descriptive statistics

North	dbh	South	dbh_1
n	23.5	s	17.2
n	43.5	s	44.6
n	6.6	s	44.1
n	11.5	s	35.5
n	17.2	s	51
n	38.7	s	21.6
n	2.3	s	44.1
n	31.5	s	11.2
n	10.5	s	36
n	23.7	s	42.1
n	13.8	s	3.2
n	5.2	s	25.5
n	31.5	s	36.5
n	22.1	s	39
n	6.7	s	25.9
n	2.6	s	20.8
n	6.3	s	3.2
n	51.1	s	57.7
n	5.4	s	43.3
n	9	s	58
n	43	s	21.7
n	8.7	s	35.6
n	22.8	s	30.9
n	2.9	s	40.6
n	22.3	s	30.7
n	43.8	s	35.6
n	48.1	s	18.2
n	46.5	s	2.9
n	39.8	s	20.4
n	10.9	s	11.4

Statistics

Variable	N	N*	Mean	SE Mean	StDev	Variance	Sum	Minimum	Q1	Median	Q3
dbh	30	0	21.72	2.93	16.07	258.38	651.50	2.30	6.68	19.65	38.98
dbh_1	30	0	30.28	2.80	15.33	235.05	908.50	2.90	19.85	33.20	42.40

Variable	Maximum	Range	IQR	Mode	N for Mode	Skewness	Kurtosis
dbh	51.10	48.80	32.30	31.5	2	0.48	-1.25
dbh_1	58.00	55.10	22.55	3.2, 35.6, 44.1	2	-0.16	-0.65

Here sample size is 30 hence t-test can be used omitting the outliers and strong skewness.

Skewness for both the samples is in between -0.5 to 0.5 hence distributions are approximately symmetric hence t-test can be applied.

b)

μ₁: mean of dbh

µ₂: mean of dbh_1

Difference: μ₁ - µ₂

Equal variances are assumed for this test

Test

Null hypothesis	H₀: μ₁ - µ₂ = 0
Alternative hypothesis	H₁: μ₁ - µ₂ ≠ 0

c)

Test

T-Value	DF	P-Value
-2.11	58	0.039

d) Confidence intervals

Estimation for Difference

Difference	Pooled StDev	95% CI for Difference
-8.57	15.71	(-16.68, -0.45)

In the confidence interval 0 (zero) is not included hence we reject null hypothesis i.e., both the samples are significantly different.