Question

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?

Sample ID	Control	Fertilizer Treatment
1	0.358	0.336
2	0.338	0.438
3	0.488	0.458
4	0.374	0.797
5	0.411	0.588
6	0.442	0.326
7	0.376	0.193
8	0.334	0.297
9	0.498	0.486
10	0.403	0.203
11	0.473	0.5
12	0.238	0.214
13	0.298	0.599
14	0.222	0.2
15	0.318	0.598
16	0.337	0.454
17	0.398	0.563
18	0.35	0.692
19	0.285	0.685
20	0.383	0.493
21	0.275	0.361
22	0.332	0.206
23	0.469	0.701
24	0.497	0.255
25	0.352	0.541
26	0.296	0.49
27	0.48	0.257
28	0.266	0.526
29	0.461	0.822
30	0.26	0.63

Answer 1

Output of the test from excel

t-Test: Two-Sample Assuming Equal Variances
	Control	Fertilizer Treatment
Mean	0.367066667	0.463633333
Variance	0.006672685	0.034469551
Observations	30	30
Pooled Variance	0.020571118
Hypothesized Mean Difference	0
df	58
t Stat	-2.607617724
P(T<=t) one-tail	0.005786759
t Critical one-tail	1.671552762
P(T<=t) two-tail	0.011573518
t Critical two-tail	2.001717484

what is the calculated t-value for this data set?
t-stat = -2.6076

This is left tailed test,
t-crit = -1.6716

p-value = 0.0058

Reject H0