A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on
a very accurate scale. The results in grams were
3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477
(a) Construct a 90 percent confidence interval for the true mean weight. (b) What sample size
would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence?
(c) Discuss the factors which might cause variation in the weight of Tootsie Rolls during
manufacture. (Data are from a project by MBA student Henry Scussel.)
(a) To construct a 90 percent confidence interval for the true mean weight, we can use the formula:
Confidence interval = sample mean ± margin of error
First, let's calculate the sample mean:
(3.087 + 3.131 + 3.241 + 3.241 + 3.270 + 3.353 + 3.400 + 3.411 + 3.437 + 3.477) / 10 = 3.3177
Next, we need to calculate the standard deviation. We can use the following formula:
Standard deviation = square root of [Σ(xᵢ - x̄)² / (n - 1)]
where xᵢ is each individual measurement, x̄ is the sample mean, and n is the sample size.
Calculating the standard deviation:
√[(3.087 - 3.3177)² + (3.131 - 3.3177)² + ... + (3.477 - 3.3177)²] / (10 - 1) = 0.1101
The margin of error can be calculated using the formula:
Margin of error = (critical value) * (standard deviation / √n)
Since we want a 90 percent confidence interval, the critical value can be obtained from a t-distribution table with n-1 degrees of freedom. For a two-tailed test, the critical value is 1.833.
Now we can calculate the margin of error:
Margin of error = 1.833 * (0.1101 / √10) ≈ 0.0602
Finally, we can construct the confidence interval:
Confidence interval = 3.3177 ± 0.0602
Confidence interval ≈ (3.2575, 3.3779)
Therefore, the 90 percent confidence interval for the true mean weight of the Tootsie Rolls is approximately (3.2575, 3.3779) grams.
(b) To determine the sample size necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence, we can use the following formula:
Sample size = (Z² * σ²) / E²
Where:
Z is the z-value corresponding to the desired confidence level (Z = 1.645 for 90% confidence),
σ is the estimated standard deviation (which we calculated above as 0.1101 grams),
E is the desired margin of error (± 0.03 grams).
Plugging in the values, we get:
Sample size = (1.645² * 0.1101²) / (0.03²)
Sample size ≈ 20.454
Therefore, a sample size of approximately 21 would be necessary to estimate the true weight with an error of ± 0.03 grams and 90 percent confidence.
(c) Factors that might cause variation in the weight of Tootsie Rolls during manufacture include:
1. Manufacturing process: The consistency of the manufacturing process, including temperature control, ingredient measurements, and mixing techniques, can affect the weight of the Tootsie Rolls.
2. Ingredient variation: Variations in the ingredients used, such as the quality and moisture content of the ingredients, can result in differences in weight.
3. Production equipment: The calibration and maintenance of the production equipment, such as the scales used for measurement, can impact the accuracy and consistency of the Tootsie Roll weights.
4. Human error: Human involvement in the production process, such as manual handling and packaging, can introduce variability in the weight of the Tootsie Rolls.
5. Environmental factors: Factors like humidity and air circulation can affect the moisture content and weight of the Tootsie Rolls.
By identifying and controlling these factors, manufacturers can aim to reduce variation in the weight of Tootsie Rolls and ensure consistent quality.
To construct a confidence interval for the true mean weight of the Tootsie Rolls, we can use the given sample data. Here are the steps:
Step 1: Calculate the sample mean.
Add up all the weights in the sample and divide by the sample size (which is 10 in this case):
3.087 + 3.131 + 3.241 + 3.241 + 3.270 + 3.353 + 3.400 + 3.411 + 3.437 + 3.477 = 33.038.
Sample mean = 33.038/10 = 3.304 grams.
Step 2: Calculate the sample standard deviation.
To do this, we first need to find the sample variance.
For each weight, subtract the sample mean and then square the result:
(3.087 - 3.304)^2 + (3.131 - 3.304)^2 + (3.241 - 3.304)^2 + (3.241 - 3.304)^2 + (3.270 - 3.304)^2 + (3.353 - 3.304)^2 + (3.400 - 3.304)^2 + (3.411 - 3.304)^2 + (3.437 - 3.304)^2 + (3.477 - 3.304)^2 = 0.1906.
Sample variance = 0.1906/9 = 0.0212.
Next, take the square root of the sample variance to find the sample standard deviation:
Sample standard deviation = √(0.0212) = 0.1455 grams.
Step 3: Determine the critical value.
For a 90 percent confidence interval, we need to find the critical value from the t-distribution with n-1 degrees of freedom (where n is the sample size). Since the sample size in this case is 10, we have 10-1=9 degrees of freedom.
By referring to the t-distribution table or using statistical software, we find that the critical value for a 90 percent confidence interval with 9 degrees of freedom is approximately 1.833.
Step 4: Calculate the margin of error.
The margin of error is the product of the critical value and the standard error.
Standard error = sample standard deviation / √(sample size) = 0.1455 / √(10) = 0.0461 grams.
Margin of error = critical value * standard error = 1.833 * 0.0461 = 0.0844 grams.
Step 5: Construct the confidence interval.
The confidence interval can be calculated by subtracting and adding the margin of error to the sample mean.
Lower limit = sample mean - margin of error = 3.304 - 0.0844 = 3.2196 grams.
Upper limit = sample mean + margin of error = 3.304 + 0.0844 = 3.3884 grams.
Therefore, the 90 percent confidence interval for the true mean weight of the Tootsie Rolls is (3.2196 grams, 3.3884 grams).
Now, addressing the second part of the question, to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence, we need to determine the required sample size. Here's how:
The formula to calculate the required sample size is:
n = (Z * σ / E)^2
Where:
n = sample size
Z = critical value (from the normal distribution, corresponding to the desired confidence level)
σ = standard deviation
E = desired margin of error
In our case:
Z = 1.833 (for the 90 percent confidence level)
σ = sample standard deviation = 0.1455 grams
E = 0.03 grams
Substituting these values into the formula:
n = (1.833 * 0.1455 / 0.03)^2
n ≈ 32.67
Since the sample size must be a whole number, we would need a sample size of at least 33 to estimate the true weight with an error of ± 0.03 grams and 90 percent confidence.
Finally, let's discuss the factors that might cause variation in the weight of Tootsie Rolls during manufacture. Some possible factors include:
1. Variation in ingredients: The ingredients mixed to make Tootsie Rolls might have natural variations in their weights or proportions, leading to slight differences in the weight of individual Tootsie Rolls.
2. Manufacturing process: Different machines or steps involved in the manufacturing process can introduce variability in the weight.
3. Temperature and humidity: These environmental factors can affect the moisture content of Tootsie Rolls, leading to variation in weight.
4. Human error: Factors like human handling, measurement errors, or inconsistencies in packaging can contribute to weight variation.
5. Batch-to-batch variation: Different batches of Tootsie Rolls may not be identical due to factors such as vendor variations, variations in raw materials, or changes in the manufacturing process.
Understanding and controlling these factors is important to ensure consistent quality in the production of Tootsie Rolls.