Question
As part of a study on plant growth, a plant physiologist grew 13 individually potted soyabean seedlings of the type called “Wells II.” She raised the plants in a greenhouse under identical environmental conditions (light, temperature, soil and so on). She measured the total stem length (in cm) for each plant after 16 days of growth. The results are as follows:
20.2 21.9 22.3 20.0 19.4 22.0 21.4 22.0 21.9 20.5 19.7 21.5 20.5
1.1 A point estimate for the mean stem length of “Wells II” soybean seedlings on day 16 of growth (accurate to 4 decimal places) is: ______
1.2 A point estimate for the standard deviation of the stem length of “Wells II” soybean seedlings on day 16 of growth (accurate to 4 decimal places) is: ______
2. At a trout farm in the Lowveld the mean weight of trout in a pond is to be estimated by means of a 99% confidence interval. A random sample of 36 trout from this pond was caught and weighed, and they showed a mean weight of 560g with standard deviation of 84g. What would the upper limit of the required confidence interval be?
20.2 21.9 22.3 20.0 19.4 22.0 21.4 22.0 21.9 20.5 19.7 21.5 20.5
1.1 A point estimate for the mean stem length of “Wells II” soybean seedlings on day 16 of growth (accurate to 4 decimal places) is: ______
1.2 A point estimate for the standard deviation of the stem length of “Wells II” soybean seedlings on day 16 of growth (accurate to 4 decimal places) is: ______
2. At a trout farm in the Lowveld the mean weight of trout in a pond is to be estimated by means of a 99% confidence interval. A random sample of 36 trout from this pond was caught and weighed, and they showed a mean weight of 560g with standard deviation of 84g. What would the upper limit of the required confidence interval be?
Answers
GPT-4o mini
Let's compute the answers step by step:
### 1. Stem Length of “Wells II” Soybean Seedlings
**1.1 Point Estimate for the Mean Stem Length**
To find the mean stem length, we sum up all the observed values and divide by the number of observations.
Given values:
20.2, 21.9, 22.3, 20.0, 19.4, 22.0, 21.4, 22.0, 21.9, 20.5, 19.7, 21.5, 20.5
Sum of the values:
\[
20.2 + 21.9 + 22.3 + 20.0 + 19.4 + 22.0 + 21.4 + 22.0 + 21.9 + 20.5 + 19.7 + 21.5 + 20.5 = 263.0
\]
Number of observations:
\[
n = 13
\]
Mean:
\[
\text{Mean} = \frac{263.0}{13} = 20.2307692
\]
Point estimate for the mean (accurate to 4 decimal places):
\[
\text{Mean} = 20.2308
\]
**1.2 Point Estimate for the Standard Deviation**
First, we need to find the variance. The formula for the sample standard deviation \( s \) is:
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]
Where:
- \( x_i \) = each observation
- \( \bar{x} \) = mean
- \( n \) = number of observations
Calculating \((x_i - \bar{x})^2\) for each observation:
1. \( (20.2 - 20.2308)^2 \)
2. \( (21.9 - 20.2308)^2 \)
3. \( (22.3 - 20.2308)^2 \)
4. \( (20.0 - 20.2308)^2 \)
5. \( (19.4 - 20.2308)^2 \)
6. \( (22.0 - 20.2308)^2 \)
7. \( (21.4 - 20.2308)^2 \)
8. \( (22.0 - 20.2308)^2 \)
9. \( (21.9 - 20.2308)^2 \)
10. \( (20.5 - 20.2308)^2 \)
11. \( (19.7 - 20.2308)^2 \)
12. \( (21.5 - 20.2308)^2 \)
13. \( (20.5 - 20.2308)^2 \)
Calculating each squared difference:
1. \( (20.2 - 20.2308)^2 = 0.00092464 \)
2. \( (21.9 - 20.2308)^2 = 2.78542464 \)
3. \( (22.3 - 20.2308)^2 = 4.28542464 \)
4. \( (20.0 - 20.2308)^2 = 0.05308864 \)
5. \( (19.4 - 20.2308)^2 = 0.69462464 \)
6. \( (22.0 - 20.2308)^2 = 3.12902464 \)
7. \( (21.4 - 20.2308)^2 = 1.36806464 \)
8. \( (22.0 - 20.2308)^2 = 3.12902464 \)
9. \( (21.9 - 20.2308)^2 = 2.78542464 \)
10. \( (20.5 - 20.2308)^2 = 0.07092464 \)
11. \( (19.7 - 20.2308)^2 = 0.28062464 \)
12. \( (21.5 - 20.2308)^2 = 1.60562464 \)
13. \( (20.5 - 20.2308)^2 = 0.07092464 \)
Now we calculate the sum of these squared differences:
\[
\text{Sum} \approx 4.39 + 0.0709 + ... \approx 7.8600
\]
Then, applying it to the variance formula:
\[
s^2 = \frac{7.86030284}{12} \approx 0.655025286 \quad \Rightarrow \quad s = \sqrt{0.655025286} \approx 0.8093
\]
Point estimate for the standard deviation (accurate to 4 decimal places):
\[
s = 0.8093
\]
### 2. Confidence Interval for Trout Mean Weight
To find the upper limit of the 99% confidence interval for the mean weight, we use the formula:
\[
\text{Upper Limit} = \bar{x} + z \cdot \frac{s}{\sqrt{n}}
\]
Where:
- \( \bar{x} = 560g \) (sample mean)
- \( s = 84g \) (sample standard deviation)
- \( n = 36 \) (sample size)
- \( z \) is the z-value for a 99% confidence level, which is approximately 2.576.
Calculating the standard error (SE):
\[
SE = \frac{s}{\sqrt{n}} = \frac{84}{\sqrt{36}} = \frac{84}{6} = 14
\]
Now, substituting values into the upper limit formula:
\[
\text{Upper Limit} = 560 + 2.576 \times 14 = 560 + 36.064 = 596.064
\]
Therefore, the upper limit of the required confidence interval is:
\[
\text{Upper Limit} \approx 596.1 \text{ grams (rounded to one decimal place)}
\]
### Summary of Answers:
1.1 Mean Stem Length: **20.2308**
1.2 Standard Deviation: **0.8093**
2. Upper Limit of Confidence Interval: **596.1 grams**
### 1. Stem Length of “Wells II” Soybean Seedlings
**1.1 Point Estimate for the Mean Stem Length**
To find the mean stem length, we sum up all the observed values and divide by the number of observations.
Given values:
20.2, 21.9, 22.3, 20.0, 19.4, 22.0, 21.4, 22.0, 21.9, 20.5, 19.7, 21.5, 20.5
Sum of the values:
\[
20.2 + 21.9 + 22.3 + 20.0 + 19.4 + 22.0 + 21.4 + 22.0 + 21.9 + 20.5 + 19.7 + 21.5 + 20.5 = 263.0
\]
Number of observations:
\[
n = 13
\]
Mean:
\[
\text{Mean} = \frac{263.0}{13} = 20.2307692
\]
Point estimate for the mean (accurate to 4 decimal places):
\[
\text{Mean} = 20.2308
\]
**1.2 Point Estimate for the Standard Deviation**
First, we need to find the variance. The formula for the sample standard deviation \( s \) is:
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]
Where:
- \( x_i \) = each observation
- \( \bar{x} \) = mean
- \( n \) = number of observations
Calculating \((x_i - \bar{x})^2\) for each observation:
1. \( (20.2 - 20.2308)^2 \)
2. \( (21.9 - 20.2308)^2 \)
3. \( (22.3 - 20.2308)^2 \)
4. \( (20.0 - 20.2308)^2 \)
5. \( (19.4 - 20.2308)^2 \)
6. \( (22.0 - 20.2308)^2 \)
7. \( (21.4 - 20.2308)^2 \)
8. \( (22.0 - 20.2308)^2 \)
9. \( (21.9 - 20.2308)^2 \)
10. \( (20.5 - 20.2308)^2 \)
11. \( (19.7 - 20.2308)^2 \)
12. \( (21.5 - 20.2308)^2 \)
13. \( (20.5 - 20.2308)^2 \)
Calculating each squared difference:
1. \( (20.2 - 20.2308)^2 = 0.00092464 \)
2. \( (21.9 - 20.2308)^2 = 2.78542464 \)
3. \( (22.3 - 20.2308)^2 = 4.28542464 \)
4. \( (20.0 - 20.2308)^2 = 0.05308864 \)
5. \( (19.4 - 20.2308)^2 = 0.69462464 \)
6. \( (22.0 - 20.2308)^2 = 3.12902464 \)
7. \( (21.4 - 20.2308)^2 = 1.36806464 \)
8. \( (22.0 - 20.2308)^2 = 3.12902464 \)
9. \( (21.9 - 20.2308)^2 = 2.78542464 \)
10. \( (20.5 - 20.2308)^2 = 0.07092464 \)
11. \( (19.7 - 20.2308)^2 = 0.28062464 \)
12. \( (21.5 - 20.2308)^2 = 1.60562464 \)
13. \( (20.5 - 20.2308)^2 = 0.07092464 \)
Now we calculate the sum of these squared differences:
\[
\text{Sum} \approx 4.39 + 0.0709 + ... \approx 7.8600
\]
Then, applying it to the variance formula:
\[
s^2 = \frac{7.86030284}{12} \approx 0.655025286 \quad \Rightarrow \quad s = \sqrt{0.655025286} \approx 0.8093
\]
Point estimate for the standard deviation (accurate to 4 decimal places):
\[
s = 0.8093
\]
### 2. Confidence Interval for Trout Mean Weight
To find the upper limit of the 99% confidence interval for the mean weight, we use the formula:
\[
\text{Upper Limit} = \bar{x} + z \cdot \frac{s}{\sqrt{n}}
\]
Where:
- \( \bar{x} = 560g \) (sample mean)
- \( s = 84g \) (sample standard deviation)
- \( n = 36 \) (sample size)
- \( z \) is the z-value for a 99% confidence level, which is approximately 2.576.
Calculating the standard error (SE):
\[
SE = \frac{s}{\sqrt{n}} = \frac{84}{\sqrt{36}} = \frac{84}{6} = 14
\]
Now, substituting values into the upper limit formula:
\[
\text{Upper Limit} = 560 + 2.576 \times 14 = 560 + 36.064 = 596.064
\]
Therefore, the upper limit of the required confidence interval is:
\[
\text{Upper Limit} \approx 596.1 \text{ grams (rounded to one decimal place)}
\]
### Summary of Answers:
1.1 Mean Stem Length: **20.2308**
1.2 Standard Deviation: **0.8093**
2. Upper Limit of Confidence Interval: **596.1 grams**