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The reason sample variance has a divisor of n−1n-1 rather than nn is that it ma...
Sep 25, 2024
Solution by Steps
step 1
The sample variance is calculated using the formula: s2=1n−1∑i=1n(xi−xˉ)2 s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 where n n is the sample size and xˉ \bar{x} is the sample mean
step 2
The divisor n−1 n-1 is used instead of n n to correct for bias in the estimation of the population variance. This is known as Bessel's correction
step 3
Using n−1 n-1 provides an unbiased estimator of the population variance, meaning that the expected value of the sample variance equals the true population variance
Answer
True
Key Concept
Sample variance and Bessel's correction
Explanation
Using n−1 n-1 instead of n n in the sample variance formula corrects bias, making it an unbiased estimator of the population variance.
Solution by Steps
step 1
The Central Limit Theorem states that if the sampled population is normally distributed, then the sampling distribution of the sample mean xˉ \bar{x} will also be normally distributed, regardless of the sample size
step 2
Since the question specifies that the sampled population is "exactly normally distributed," the sampling distribution of xˉ \bar{x} will indeed be normal for any sample size n n
step 3
Therefore, the statement in the question is true
Answer
A
Key Concept
Sampling Distribution of the Sample Mean
Explanation
If the population is normally distributed, the sampling distribution of the sample mean is also normally distributed, regardless of sample size.
A minimum-variance unbiased point estimate has a variance that is as small as or smaller than the variances of any other unbiased point estimate. 答案选项组
Solution by Steps
step 1
A minimum-variance unbiased estimator (MVUE) is defined as an unbiased estimator that has the lowest variance among all unbiased estimators for a parameter
step 2
This means that if you have multiple unbiased estimators for the same parameter, the MVUE will have a variance that is less than or equal to the variances of those other estimators
step 3
Therefore, the statement in the question is true, as it correctly describes the property of a minimum-variance unbiased point estimate
Answer
True
Key Concept
Minimum-variance unbiased estimator (MVUE) is the estimator with the smallest variance among all unbiased estimators.
Explanation
The statement accurately reflects the definition of MVUE, confirming that it has a variance that is as small as or smaller than any other unbiased point estimate.
The sample standard deviation s is an unbiased estimator of the population standard deviation σ.
Solution by Steps
step 1
The sample standard deviation s s is calculated using the formula: s=1n−1∑i=1n(xi−xˉ)2 s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} where n n is the sample size and xˉ \bar{x} is the sample mean
step 2
The population standard deviation σ \sigma is defined as: σ=1N∑i=1N(xi−μ)2 \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} where N N is the population size and μ \mu is the population mean
step 3
The sample standard deviation s s is not an unbiased estimator of the population standard deviation σ \sigma because while s2 s^2 (the sample variance) is an unbiased estimator of σ2 \sigma^2 , the square root transformation does not preserve this property
Answer
False
Key Concept
The sample standard deviation is not an unbiased estimator of the population standard deviation.
Explanation
While the sample variance is unbiased, the sample standard deviation does not maintain this property due to the non-linear transformation involved in taking the square root.
For any sampled population, the population of all sample means is approximately normally distributed.
Solution by Steps
step 1
According to the Central Limit Theorem, the sampling distribution of the sample mean xˉ \bar{x} will be approximately normally distributed if the sample size is sufficiently large, regardless of the population's distribution
step 2
However, if the population is not normally distributed, the approximation improves with larger sample sizes. For any sampled population, the distribution of sample means approaches normality as the sample size increases
step 3
Therefore, the statement that "the population of all sample means is approximately normally distributed" is true when considering large enough sample sizes, but it may not hold for small samples from non-normal populations
Answer
True
Key Concept
Central Limit Theorem
Explanation
The Central Limit Theorem states that the distribution of sample means approaches normality as the sample size increases, regardless of the population's distribution.
Consider two population distributions labeled A and B. Distribution A is highly skewed and nonnormal, while distribution B is slightly skewed and near normal. In order for the sampling distributions of A and B to achieve the same degree of normality, 答案选项组 populations A and B will require the same sample size. population A will require a larger sample size. None of these answers is correct. population B will require a larger sample size.
Solution by Steps
step 1
According to the Central Limit Theorem, the sampling distribution of the sample mean will approach normality as the sample size increases, especially for non-normal populations
step 2
For a highly skewed distribution (like population A), a larger sample size is generally required to achieve a similar degree of normality compared to a slightly skewed distribution (like population B)
step 3
Therefore, population A will require a larger sample size than population B to achieve the same degree of normality in their sampling distributions
Answer
B
Key Concept
Sample size and normality in sampling distributions
Explanation
Larger sample sizes are needed for highly skewed distributions to achieve normality in sampling distributions compared to less skewed distributions.
It has been reported that the average time to download the home page from a government website was .9 seconds. Suppose that the download times were normally distributed with a standard deviation of .3 seconds. If random samples of 36 download times are selected, what is the probability that the sample mean will be less than .84 seconds?
Solution by Steps
step 1
We need to find the probability that the sample mean xˉ \bar{x} is less than 0.84 seconds. Given that the population mean μ=0.9 \mu = 0.9 seconds and the standard deviation σ=0.3 \sigma = 0.3 seconds, we can calculate the standard error (SE) of the sample mean using the formula: SE=σn=0.336=0.05 SE = \frac{\sigma}{\sqrt{n}} = \frac{0.3}{\sqrt{36}} = 0.05
step 2
Next, we standardize the sample mean using the z-score formula: z=xˉ−μSE=0.84−0.90.05=−1.2 z = \frac{\bar{x} - \mu}{SE} = \frac{0.84 - 0.9}{0.05} = -1.2
step 3
Now, we look up the z-score of -1.2 in the standard normal distribution table, which gives us the probability that xˉ \bar{x} is less than 0.84 seconds. The corresponding probability is approximately 0.1151
Answer
The probability that the sample mean will be less than 0.84 seconds is approximately 0.1151.
Key Concept
The Central Limit Theorem allows us to use the normal distribution to find probabilities related to sample means.
Explanation
By calculating the standard error and using the z-score, we can determine the probability of the sample mean being below a certain value.
A manufacturing company measures the weight of boxes before shipping them to the customers. If the box weights have a population mean and standard deviation of 90 lb and 24 lb, respectively, then based on a sample size of 36 boxes, what is the probability that the average weight of the boxes will exceed 94 lb?
Solution by Steps
step 1
We need to find the probability that the sample mean xˉ \bar{x} exceeds 94 lb. Given that the population mean μ=90 \mu = 90 lb and the standard deviation σ=24 \sigma = 24 lb, we can calculate the standard error (SE) of the sample mean using the formula: SE=σn=2436=4 SE = \frac{\sigma}{\sqrt{n}} = \frac{24}{\sqrt{36}} = 4
step 2
Next, we standardize the sample mean using the z-score formula: z=xˉ−μSE=94−904=1 z = \frac{\bar{x} - \mu}{SE} = \frac{94 - 90}{4} = 1
step 3
Now, we find the probability that z z is greater than 1. Using the standard normal distribution table, we find P(Z > 1) . The cumulative probability for z=1 z = 1 is approximately 0.8413, so: P(Z > 1) = 1 - P(Z \leq 1) = 1 - 0.8413 = 0.1587
Answer
The probability that the average weight of the boxes will exceed 94 lb is approximately 0.1587.
Key Concept
The Central Limit Theorem allows us to use the normal distribution to find probabilities related to sample means.
Explanation
The calculated probability indicates that there is about a 15.87% chance that the average weight of the boxes will exceed 94 lb based on the given population parameters.
The diameter of small Nerf balls manufactured overseas is expected to be approximately normally distributed with a mean of 5.2 inches and a standard deviation of .08 inches. Suppose a random sample of 20 balls is selected. Calculate the standard deviation of the sampling distribution of the sample mean.
Solution by Steps
step 1
To calculate the standard deviation of the sampling distribution of the sample mean, we use the formula for the standard error (SE): SE=σn SE = \frac{\sigma}{\sqrt{n}} where σ \sigma is the population standard deviation and n n is the sample size
step 2
Given that the population standard deviation σ=0.08 \sigma = 0.08 inches and the sample size n=20 n = 20 , we can substitute these values into the formula: SE=0.0820 SE = \frac{0.08}{\sqrt{20}}
step 3
Calculating the standard error: SE=0.0820≈0.084.472≈0.0179 SE = \frac{0.08}{\sqrt{20}} \approx \frac{0.08}{4.472} \approx 0.0179
Answer
The standard deviation of the sampling distribution of the sample mean is approximately 0.0179 inches.
Key Concept
The standard error measures the variability of the sample mean from the population mean.
Explanation
A smaller standard error indicates that the sample mean is likely to be closer to the population mean, reflecting less variability in the sample means.
Packages of sugar bags for Sweeter Sugar Inc. have an average weight of 16 ounces and a standard deviation of .3 ounces. The weights of the sugar packages are normally distributed. What is the probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces?
Solution by Steps
step 1
We need to find the probability that the average weight of 9 randomly selected sugar packages exceeds 16.025 ounces. Given that the population mean μ=16 \mu = 16 ounces and the population standard deviation σ=0.3 \sigma = 0.3 ounces, we first calculate the standard error (SE) of the sample mean. The formula for SE is: SE=σn SE = \frac{\sigma}{\sqrt{n}} where n=9 n = 9
step 2
Calculating the standard error: SE=0.39=0.33=0.1 SE = \frac{0.3}{\sqrt{9}} = \frac{0.3}{3} = 0.1
step 3
Next, we standardize the sample mean using the z-score formula: z=xˉ−μSE z = \frac{\bar{x} - \mu}{SE} where xˉ=16.025 \bar{x} = 16.025 . Thus, we calculate: z=16.025−160.1=0.0250.1=0.25 z = \frac{16.025 - 16}{0.1} = \frac{0.025}{0.1} = 0.25
step 4
Now, we find the probability that the z-score is greater than 0.25. Using the standard normal distribution table, we find P(Z > 0.25) . The cumulative probability for z=0.25 z = 0.25 is approximately 0.5987, so: P(Z > 0.25) = 1 - P(Z \leq 0.25) = 1 - 0.5987 = 0.4013
Answer
The probability that the average weight of 9 randomly selected sugar packages exceeds 16.025 ounces is approximately 0.4013.
Key Concept
This problem illustrates the use of the Central Limit Theorem and the calculation of probabilities for sample means.
Explanation
The solution involves calculating the standard error, standardizing the sample mean to find the z-score, and then using the standard normal distribution to find the probability
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