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 |
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 [X] hypothesis states that the average size of the carrots from each treatment are the same, whereas the [X] hypothesis states that the fertilized carrots are larger in size.
To analyze this data set, the scientist should use a [X]-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?
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.
3. What is the appropriate p value for the t-test?
4. 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...
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 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...
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1 = 4, s1 = .20, n1 = 15, x⎯⎯2 = 4.25, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1 = 4, s1 = .20, n1 = 15, x⎯⎯2 = 4.25, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1x1 = 4.75, s1 = .20, n1 = 15, x⎯⎯2x2 = 5.18, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1x1 = 4.75, s1 = .20, n1 = 15, x⎯⎯2x2 = 5.18, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. (a-1) Comparison of GPA for randomly chosen college juniors and seniors: x⎯⎯1x1 = 4.75, s1 = .20, n1 = 15, x⎯⎯2x2 = 5.18, s2 = .30, n2 = 15, α = .025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick"...
The following are six observations collected from treatment 1,
ten observations collected from treatment, and eight observations
collected from treatment 3. Test the hypothesis that the treatment
means are equal at the 0.05 significance level
??⭐️
The following are six observations collected from treatment 1, ten observations collected from treatment 2, and eight observations collected from treatment 3. Test the hypothesis that the treatment means are equal at the 0.05 significance level Treatment 1 Treatment 2 Treatment 3 6 2...
Treatment 1 Treatment 2 Treatment 3 0 1 6 1 4 5 0 1 6 3 2 3 T = 4 T = 8 T = 20 SS = 6 SS = 6 SS = 6 N = 12 G = 32 ƩX2= 138 1a. Conduct a single-factor independent-measures ANOVA to test the hypothesis that there are significant differences in the mean scores among the three treatment conditions. Use α = .01. The alternative hypothesis is Group of answer choices...
To test the efficacy of three different brands of fertilizer, you grow seedlings on each of the brands, plus a control with no fertilizer, for a total of four treatments. At the end of two months, you weigh each seedling. Here are the data, in grams: fertilizer A fertilizer B fertilizer C control 12 15 11 9 15 16 13 11 11 14 17 10 14 18 14 13 12 18 12 9 14 13 14 11 The question of...