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 null hypothesis states that the average size of the carrots from each treatment are the same, whereas the alternate hypothesis states that the fertilized carrots are larger in size.

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

1.) 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 4 decimal places.
• Report your answer as the absolute value of your calculated t statistic.

2.) 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 4 decimal places.

3.) 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

4.) Would you reject or fail to reject the null hypothesis?

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

Answer 1

The Excel output is:

Control	Nitrogen Treatment
0.33683	0.52527	mean
0.08993	0.19635	std. dev.
30	30	n
	58	df
	-0.188433	difference (Control - Nitrogen Treatment)
	0.023321	pooled variance
	0.152711	pooled std. dev.
	0.039430	standard error of difference
	0	hypothesized difference
	1.67155	t Critical one-tail
	-4.7790	t
	6.22E-06	p-value (one-tailed, lower)

1.) 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?

-4.7790

2.) 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.

1.6716

3.) What is the appropriate p value for the t-test?

6.22E-06

4.) Would you reject or fail to reject the null hypothesis?

Reject the null hypothesis