Question
In an agricultural experiment, the effects of two fertilizers on the production of oranges were measured. Thirteen randomly selected plots of land were treated with fertilizer A, and 8 randomly selected plots were treated with fertilizer B. The number of pounds of harvested fruit was measured from each plot. Following are the results.
Fertilizer A:
445
510
464
472
441
480
403
460
448
457
437
505
417
Fertilizer B
398
380
368
393
424
387
378
415
Construct a 95% confidence interval for the difference between the mean yields for the two types of fertilizer. Let uv1 denote the mean yield for fertilizer A. Round answers to one decimal point
Fertilizer A:
445
510
464
472
441
480
403
460
448
457
437
505
417
Fertilizer B
398
380
368
393
424
387
378
415
Construct a 95% confidence interval for the difference between the mean yields for the two types of fertilizer. Let uv1 denote the mean yield for fertilizer A. Round answers to one decimal point
Answers
Answered by
GPT-4o mini
To construct a 95% confidence interval for the difference between the mean yields of two fertilizers, we will follow these steps:
1. **Calculate the sample means** and **sample standard deviations** for both groups (Fertilizer A and Fertilizer B).
2. **Determine the standard error** of the difference between the two means.
3. **Find the critical value** from the t-distribution.
4. **Compute the confidence interval** for the difference between the means.
### Step 1: Calculate Sample Means and Standard Deviations
**Fertilizer A:**
- Data: 445, 510, 464, 472, 441, 480, 403, 460, 448, 457, 437, 505, 417
- Sample size \( n_1 = 13 \)
\[
\bar{x}_1 = \frac{445 + 510 + 464 + 472 + 441 + 480 + 403 + 460 + 448 + 457 + 437 + 505 + 417}{13}
\]
Calculating the sum:
\[
\text{Sum} = 445 + 510 + 464 + 472 + 441 + 480 + 403 + 460 + 448 + 457 + 437 + 505 + 417 = 5831
\]
Calculating the mean:
\[
\bar{x}_1 = \frac{5831}{13} \approx 448.5385 \approx 448.5
\]
Next, we calculate the sample standard deviation \( s_1 \):
\[
s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1}
\]
Calculating \( (x_i - \bar{x}_1)^2 \) for each \( x_i \):
\[
= (445 - 448.5)^2 + (510 - 448.5)^2 + (464 - 448.5)^2 + (472 - 448.5)^2 + (441 - 448.5)^2 + (480 - 448.5)^2 + (403 - 448.5)^2 + (460 - 448.5)^2 + (448 - 448.5)^2 + (457 - 448.5)^2 + (437 - 448.5)^2 + (505 - 448.5)^2 + (417 - 448.5)^2
\]
Calculating those squared differences gives:
\[
= (3.5^2) + (61.5^2) + (15.5^2) + (23.5^2) + (7.5^2) + (31.5^2) + (45.5^2) + (11.5^2) + (0.5^2) + (8.5^2) + (11.5^2) + (56.5^2) + (31.5^2)
\]
Calculating each term gives approximately \( 12.25 + 3782.25 + 240.25 + 552.25 + 56.25 + 992.25 + 2070.25 + 132.25 + 0.25 + 72.25 + 132.25 + 3192.25 + 992.25 \).
Summing these values gives \( 8867.5 \).
Now calculate \( s_1 \):
\[
s_1^2 = \frac{8867.5}{12} \approx 738.9583 \quad \Rightarrow \quad s_1 \approx 27.2
\]
**Fertilizer B:**
- Data: 398, 380, 368, 393, 424, 387, 378, 415
- Sample size \( n_2 = 8 \)
\[
\bar{x}_2 = \frac{398 + 380 + 368 + 393 + 424 + 387 + 378 + 415}{8}
\]
Calculating the sum:
\[
\text{Sum} = 398 + 380 + 368 + 393 + 424 + 387 + 378 + 415 = 3143
\]
Calculating the mean:
\[
\bar{x}_2 = \frac{3143}{8} = 392.875 \approx 392.9
\]
Next, we calculate the sample standard deviation \( s_2 \):
\[
s_2^2 = \frac{\sum (x_i - \bar{x}_2)^2}{n_2 - 1}
\]
Calculating \( (x_i - \bar{x}_2)^2 \):
\[
= (398 - 392.9)^2 + (380 - 392.9)^2 + (368 - 392.9)^2 + (393 - 392.9)^2 + (424 - 392.9)^2 + (387 - 392.9)^2 + (378 - 392.9)^2 + (415 - 392.9)^2
\]
Each squared difference:
\[
= (5.1^2) + (-12.9^2) + (-24.9^2) + (0.1^2) + (31.1^2) + (-5.9^2) + (-14.9^2) + (22.1^2)
\]
Calculating gives \( 26.01 + 166.41 + 620.01 + 0.01 + 966.41 + 34.81 + 222.01 + 488.41 = 2512.5 \).
Now calculate \( s_2 \):
\[
s_2^2 = \frac{2512.5}{7} \approx 358.9286 \quad \Rightarrow \quad s_2 \approx 18.9
\]
### Step 2: Determine Standard Error of Difference
The standard error of the difference between the two means is given by:
\[
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{738.9583}{13} + \frac{358.9286}{8}}
\]
Calculating, we find:
\[
SE = \sqrt{56.0673 + 44.8661} \approx \sqrt{100.9334} \approx 10.0
\]
### Step 3: Find the Critical Value
Degrees of freedom can be approximated using the following formula:
\[
df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}
\]
Calculating \( df \):
\[
= \frac{(56.0673 + 44.8661)^2}{\frac{(56.0673)^2}{12} + \frac{(44.8661)^2}{7}} = \frac{100.9334^2}{\frac{3147.4}{12} + \frac{2017.7}{7}}
\]
Calculating this gives degrees of freedom approximately 17.5. For practical purposes, we round it down to 17. We consult a t-distribution table or calculator for \( t_{0.025, 17} \approx 2.110 \).
### Step 4: Calculate the Confidence Interval
The difference in means:
\[
\bar{x}_1 - \bar{x}_2 \approx 448.5 - 392.9 = 55.6
\]
Thus, the confidence interval is:
\[
(\bar{x}_1 - \bar{x}_2) \pm t \cdot SE
\]
\[
= 55.6 \pm 2.110 \cdot 10.0
\]
\[
= 55.6 \pm 21.1
\]
Calculating the interval:
\[
= (34.5, 76.7)
\]
### Conclusion
The 95% confidence interval for the difference between the mean yields for the two types of fertilizer is approximately \( (34.5, 76.7) \).
1. **Calculate the sample means** and **sample standard deviations** for both groups (Fertilizer A and Fertilizer B).
2. **Determine the standard error** of the difference between the two means.
3. **Find the critical value** from the t-distribution.
4. **Compute the confidence interval** for the difference between the means.
### Step 1: Calculate Sample Means and Standard Deviations
**Fertilizer A:**
- Data: 445, 510, 464, 472, 441, 480, 403, 460, 448, 457, 437, 505, 417
- Sample size \( n_1 = 13 \)
\[
\bar{x}_1 = \frac{445 + 510 + 464 + 472 + 441 + 480 + 403 + 460 + 448 + 457 + 437 + 505 + 417}{13}
\]
Calculating the sum:
\[
\text{Sum} = 445 + 510 + 464 + 472 + 441 + 480 + 403 + 460 + 448 + 457 + 437 + 505 + 417 = 5831
\]
Calculating the mean:
\[
\bar{x}_1 = \frac{5831}{13} \approx 448.5385 \approx 448.5
\]
Next, we calculate the sample standard deviation \( s_1 \):
\[
s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1}
\]
Calculating \( (x_i - \bar{x}_1)^2 \) for each \( x_i \):
\[
= (445 - 448.5)^2 + (510 - 448.5)^2 + (464 - 448.5)^2 + (472 - 448.5)^2 + (441 - 448.5)^2 + (480 - 448.5)^2 + (403 - 448.5)^2 + (460 - 448.5)^2 + (448 - 448.5)^2 + (457 - 448.5)^2 + (437 - 448.5)^2 + (505 - 448.5)^2 + (417 - 448.5)^2
\]
Calculating those squared differences gives:
\[
= (3.5^2) + (61.5^2) + (15.5^2) + (23.5^2) + (7.5^2) + (31.5^2) + (45.5^2) + (11.5^2) + (0.5^2) + (8.5^2) + (11.5^2) + (56.5^2) + (31.5^2)
\]
Calculating each term gives approximately \( 12.25 + 3782.25 + 240.25 + 552.25 + 56.25 + 992.25 + 2070.25 + 132.25 + 0.25 + 72.25 + 132.25 + 3192.25 + 992.25 \).
Summing these values gives \( 8867.5 \).
Now calculate \( s_1 \):
\[
s_1^2 = \frac{8867.5}{12} \approx 738.9583 \quad \Rightarrow \quad s_1 \approx 27.2
\]
**Fertilizer B:**
- Data: 398, 380, 368, 393, 424, 387, 378, 415
- Sample size \( n_2 = 8 \)
\[
\bar{x}_2 = \frac{398 + 380 + 368 + 393 + 424 + 387 + 378 + 415}{8}
\]
Calculating the sum:
\[
\text{Sum} = 398 + 380 + 368 + 393 + 424 + 387 + 378 + 415 = 3143
\]
Calculating the mean:
\[
\bar{x}_2 = \frac{3143}{8} = 392.875 \approx 392.9
\]
Next, we calculate the sample standard deviation \( s_2 \):
\[
s_2^2 = \frac{\sum (x_i - \bar{x}_2)^2}{n_2 - 1}
\]
Calculating \( (x_i - \bar{x}_2)^2 \):
\[
= (398 - 392.9)^2 + (380 - 392.9)^2 + (368 - 392.9)^2 + (393 - 392.9)^2 + (424 - 392.9)^2 + (387 - 392.9)^2 + (378 - 392.9)^2 + (415 - 392.9)^2
\]
Each squared difference:
\[
= (5.1^2) + (-12.9^2) + (-24.9^2) + (0.1^2) + (31.1^2) + (-5.9^2) + (-14.9^2) + (22.1^2)
\]
Calculating gives \( 26.01 + 166.41 + 620.01 + 0.01 + 966.41 + 34.81 + 222.01 + 488.41 = 2512.5 \).
Now calculate \( s_2 \):
\[
s_2^2 = \frac{2512.5}{7} \approx 358.9286 \quad \Rightarrow \quad s_2 \approx 18.9
\]
### Step 2: Determine Standard Error of Difference
The standard error of the difference between the two means is given by:
\[
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{738.9583}{13} + \frac{358.9286}{8}}
\]
Calculating, we find:
\[
SE = \sqrt{56.0673 + 44.8661} \approx \sqrt{100.9334} \approx 10.0
\]
### Step 3: Find the Critical Value
Degrees of freedom can be approximated using the following formula:
\[
df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}
\]
Calculating \( df \):
\[
= \frac{(56.0673 + 44.8661)^2}{\frac{(56.0673)^2}{12} + \frac{(44.8661)^2}{7}} = \frac{100.9334^2}{\frac{3147.4}{12} + \frac{2017.7}{7}}
\]
Calculating this gives degrees of freedom approximately 17.5. For practical purposes, we round it down to 17. We consult a t-distribution table or calculator for \( t_{0.025, 17} \approx 2.110 \).
### Step 4: Calculate the Confidence Interval
The difference in means:
\[
\bar{x}_1 - \bar{x}_2 \approx 448.5 - 392.9 = 55.6
\]
Thus, the confidence interval is:
\[
(\bar{x}_1 - \bar{x}_2) \pm t \cdot SE
\]
\[
= 55.6 \pm 2.110 \cdot 10.0
\]
\[
= 55.6 \pm 21.1
\]
Calculating the interval:
\[
= (34.5, 76.7)
\]
### Conclusion
The 95% confidence interval for the difference between the mean yields for the two types of fertilizer is approximately \( (34.5, 76.7) \).
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