Question

Important instructions: For all situations requiring a hypothesis test (z test, one-samplet test, one-sample variance test or two-sample t test), you must 1. Choose the appropriate test based on the information you are given 2. State null and alternative hypotheses 3. Choose a one or two tailed test and explain why you chose that test 4. Calculate the appropriate test statistic, showing all work neatly. This includes calculations of means, standard deviations, etc. 5. Draw the appropriate conclusions (i.e., do you reject or fail to reject Ho?) To help you know for sure which questions require this, I have marked the questions requiring these steps with an asterisk (*). You may assume data are normal and have equal variances-that necessary assumptions have been fulfilled. Use your own paper and label each question clearly. You may not receive full credit if the work is disorganized or difficult to follow. 1. *(10 points) You may assume these data are normal and have equal variances. You are comparing acorn production in two different locations. Does the Mesic Soil area lead to a greater number of acorns per five square meter sample than the Sand Bottoms arca? Mesic Soil Sand Bottoms 93 95 81 05

Answer 1

We use a 2 sample t test for conducting the test

Let $\mu_1$ be the the average number of acorns in Mesic Soil.

Let $\mu_2$ be the the average number of acorns in Sand Bottom.

From the data

For: $\bar{x_1}$ = 95, s₁ = 3, n₁ = 10

For: $\bar{x_2}$ = 85.6, s₂ = 3.69, n₂ = 10

$S_p^2 = \frac{(n1-1)*s1^2+(n2-1)*s2^2}{n1+n2-2} = \frac{9*3^2+9*3.69^2}{10+10-2} = 11.3081$

The Hypothesis:

H0: $\mu_1$ = $\mu_2$

Ha: $\mu_1$ > $\mu_2$

This is a Right tailed test.

The Test Statistic:

$t = \frac{\bar{x}_1-\bar{x}_2}{\sqrt{S_p^2}(\frac{1}{n1}+\frac{1}{n2})} = \frac{95-85.6}{\sqrt{11.3081*(\frac{1}{10}+\frac{1}{10})}} = 6.25$

The p Value:    The p value (Right Tail) for t = 6.25 ,df = 18,is; p value = 0.0000

The Critical Value:   The critical value (Right tail) at $\alpha$ = 0.05 (default), df = 18,tcritical = +1.734

The Decision Rule:    If tobserved is > tcritical, Then Reject H0.

Also If the P value is < $\alpha$ , Then Reject H0

The Decision:    Since t observed (6.25) is > tcritical (1.734), We Reject H0.

Also since P value (0.0000) is < $\alpha$ (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that the mesic soil area leads to greater number of acorns per 5 square feet than Sand Bottom.

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Calculation for mean and Standard Deviation

Mean = Sum of Observations / Total Observations

Variance = SS/n - 1, where SS = Sum of squares

SS = SUM(X - Mean)².

Standard Deviation = Sqrt(Variance)

	Mesic	Sand
n	10	10
Sum	950	856
Average	95.00	85.6
SS(Sum of squares)	82	122.4
Variance = SS/n-1	9.11	13.60
Std Dev=Sqrt(Variance)	3.0	3.69

Mesic Soil						Sand Bottoms
#	X	Mean	(x - mean)2			#	X	Mean	(x - mean)2
1	95	95.000	0.000			1	93	85.600	54.760
2	93	95.000	4.000			2	83	85.600	6.760
3	96	95.000	1.000			3	81	85.600	21.160
4	92	95.000	9.000			4	85	85.600	0.360
5	98	95.000	9.000			5	85	85.600	0.360
6	91	95.000	16.000			6	88	85.600	5.760
7	95	95.000	0.000			7	83	85.600	6.760
8	100	95.000	25.000			8	82	85.600	12.960
9	98	95.000	9.000			9	89	85.600	11.560
10	92	95.000	9.000			10	87	85.600	1.960