To construct a 95% confidence interval for the difference between the mean yields of two fertilizers, we will follow these steps:
- Calculate the sample means and sample standard deviations for both groups (Fertilizer A and Fertilizer B).
- Determine the standard error of the difference between the two means.
- Find the critical value from the t-distribution.
- 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 \( (34.5, 76.7) \).