Question

Sample ID	Control	Nitrogen Treatment
1	0.331	0.419
2	0.375	0.727
3	0.385	0.766
4	0.474	0.741
5	0.223	0.182
6	0.261	0.821
7	0.4	0.251
8	0.349	0.778
9	0.203	0.641
10	0.332	0.368
11	0.231	0.732
12	0.457	0.453
13	0.216	0.45
14	0.29	0.196
15	0.218	0.325
16	0.353	0.706
17	0.285	0.5
18	0.463	0.691
19	0.371	0.691
20	0.411	0.755
21	0.407	0.35
22	0.249	0.395
23	0.298	0.519
24	0.4	0.285
25	0.361	0.423
26	0.488	0.363
27	0.338	0.737
28	0.215	0.538
29	0.486	0.626
30	0.235	0.329

A soil scientist has just developed a new type of fertilizer and she wants to determine whether it helps carrots grow larger. She sets up several pots of soil and plants one carrot seed in each pot. Fertilizer is added to half the pots. All the pots are placed in a temperature-controlled greenhouse where they receive adequate light and equal amounts of water. After two months of growth, the scientist harvests the carrots and weighs them (in kilograms). Below is a data table showing the weight of the carrots at the end of the growing period from the two treatment groups. Here is a hyperlink to the data.

When analyzing this dataset with a t-test, the  hypothesis states that the average size of the carrots from each treatment are the same, whereas the  hypothesis states that the fertilized carrots are larger in size.

To analyze this data set, the scientist should use a -tailed t-test.

After performing a t-test assuming equal variances using the data analysis add-in for MS Excel, what is the calculated t-value for this data set?

Round your answer to four decimal places.

Your answer should be a positive value.

Report the appropriate critical t value, based on your decision of a one- or two-tailed test, calculated by MS Excel using the Data Analysis add-in.

Round your answer to four decimal places.

What is the appropriate p value for the t-test?

Report your answer in exponential notation

Report your answer to 4 decimal places after converting to exponential notation

e.g.

1.1234E-01 for 0.112341

3.1234E-04 for 0.000312341

Would you reject or fail to reject the null hypothesis?

Answer 1

Solution:

We have to test the hypothesis that: the average size of the carrots from each treatment are the same, whereas the  hypothesis states that the fertilized carrots are larger in size.

That is:

We have to perform a t-test assuming equal variances using the data analysis add-in for MS Excel.

Use following steps in MS Excel:

i) Click on Data tab

ii) Select Data Analysis

iii) Select t-Test: Two-Sample Assuming Equal Variances

Under t-Test: Two-Sample Assuming Equal Variances, select Nitrogen Treatment for Variable 1 Range and Control for Variable 2 range,

Hypothesized difference = 0

Labels Click on Check Box

Click on OK

We will get output in new sheet.

Thus we get:

t-Test: Two-Sample Assuming Equal Variances
	Nitrogen Treatment	Control
Mean	0.52527	0.336833
Variance	0.03855	0.008087
Observations	30	30
Pooled Variance	0.02332
Hypothesized Mean Difference	0
df	58
t Stat	4.77895
P(T<=t) one-tail	0.00001
t Critical one-tail	1.67155
P(T<=t) two-tail	0.00001
t Critical two-tail	2.00172

Thus :

i) The calculated t-value for this data set = t = 4.7790

ii) This is one tailed test, hence t critical One tail = 1.6716

iii) the appropriate p value for the t-test = P(T<=t) one-tail) = 0.00001 = 0.0000

iv) Would you reject or fail to reject the null hypothesis?

Yes , since P-value = 0.0000 < 0.05 level of significance.

Thus we conclude that: the fertilized carrots are larger in size.