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问题 1 1 分 If the sample size n is infinitely large, then s2s^{2} is an unbiased...
Sep 25, 2024
Solution by Steps
step 2
As the sample size approaches infinity, the sample variance s2 s^{2} converges in probability to the population variance σ2 \sigma^{2} . This means that s2 s^{2} becomes an unbiased estimator of σ2 \sigma^{2} as n n increases
step 3
Therefore, the statement is true because with an infinitely large sample size, the bias of the estimator s2 s^{2} approaches zero
step 4
Thus, we conclude that the answer to the question is "True"
A
Key Concept
Unbiased Estimator
Explanation
An estimator is unbiased if its expected value equals the parameter it estimates. As sample size increases, sample variance s2 s^{2} becomes an unbiased estimator of population variance σ2 \sigma^{2} .
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 True False
Solution by Steps
step 2
By definition, a minimum-variance unbiased estimator (MVUE) is an unbiased estimator that achieves the lowest variance among all unbiased estimators for a parameter
step 3
Therefore, the statement is true because the MVUE is specifically designed to have the smallest variance compared to other unbiased estimators
step 4
Thus, we conclude that the answer to the question is "True"
A
Key Concept
Minimum-Variance Unbiased Estimator (MVUE)
Explanation
An MVUE is an unbiased estimator that has the smallest variance among all unbiased estimators for a given parameter.
The Central Limit Theorem states that as sample size increases, the sample mean distribution more closely approximates a normal distribution. 答案选项组 True False
Solution by Steps
step 2
This means that for sufficiently large sample sizes, the sample mean will be approximately normally distributed, which is a fundamental concept in statistics
step 3
Therefore, the statement in the question is true as it accurately reflects the essence of the Central Limit Theorem
A
Key Concept
Central Limit Theorem
Explanation
As sample size increases, the distribution of the sample mean approaches a normal distribution, regardless of the population's distribution.
If the sampled population is exactly normally distributed, then the sampling distribution of is also expected to be normal, regardless of the sample size. 答案选项组 True False
Solution by Steps
step 2
Since the population is normally distributed, the sampling distribution of the sample mean will also be normally distributed for any sample size n n
step 3
Therefore, the statement is true as it accurately reflects the properties of normal distributions
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.
Solution by Steps
step 2
Since the population is normally distributed, the sampling distribution of the sample mean xˉ \bar{x} will also be normally distributed for any sample size n n . This is a fundamental property of normal distributions
step 3
Therefore, the statement is true as it accurately reflects the properties of normal distributions
A
Key Concept
Normal Distribution Properties
Explanation
If the population is normally distributed, the sampling distribution of the sample mean is also normally distributed, regardless of sample size.
Solution by Steps
step 2
Since the population is normally distributed, the sampling distribution of the sample mean xˉ \bar{x} will also be normally distributed for any sample size n n
step 3
Therefore, the statement is true as it accurately reflects the properties of normal distributions
A
Key Concept
Normal Distribution Properties
Explanation
If a population is normally distributed, the sampling distribution of the sample mean is also normally distributed, regardless of sample size.
Solution by Steps
step 2
First, we calculate the standard error (SE) of the sample mean using the formula SE=σn SE = \frac{\sigma}{\sqrt{n}} . Thus, SE=20100=2 SE = \frac{20}{\sqrt{100}} = 2
step 3
Next, we standardize the values using the z-score formula: z=xˉ−μSE z = \frac{\bar{x} - \mu}{SE} . For xˉ=395.4 \bar{x} = 395.4 , z1=395.4−4002=−2.3 z_1 = \frac{395.4 - 400}{2} = -2.3 and for xˉ=404.6 \bar{x} = 404.6 , z2=404.6−4002=2.3 z_2 = \frac{404.6 - 400}{2} = 2.3
step 4
Now, we find the probabilities corresponding to these z-scores using the standard normal distribution table: P(Z < -2.3) \approx 0.0107 and P(Z < 2.3) \approx 0.9893
step 5
Finally, we calculate the probability P(395.4 < \bar{x} < 404.6) = P(Z < 2.3) - P(Z < -2.3) = 0.9893 - 0.0107 = 0.9786
A
Key Concept
Central Limit Theorem
Explanation
The Central Limit Theorem states that the distribution of the sample mean approaches a normal distribution as the sample size increases, allowing us to use z-scores to find probabilities.
Solution by Steps
step 2
Next, we standardize the value 16.025 using the Z-score formula: Z=xˉ−μSE Z = \frac{\bar{x} - \mu}{SE} . Therefore, Z=16.025−160.1=0.25 Z = \frac{16.025 - 16}{0.1} = 0.25
step 3
We now find the probability P(\bar{x} > 16.025) which is equivalent to P(Z > 0.25) . Using the standard normal distribution table, we find P(Z < 0.25) \approx 0.5987 . Thus, P(Z > 0.25) = 1 - P(Z < 0.25) = 1 - 0.5987 = 0.4013
[1] Answer
D
Key Concept
Normal Distribution and Z-scores
Explanation
The Z-score allows us to determine the probability of a sample mean exceeding a certain value in a normal distribution, using the mean and standard error.
Solution by Steps
step 2
First, we calculate the standard error (SE) of the sample mean using the formula SE=σn SE = \frac{\sigma}{\sqrt{n}} . Thus, SE=0.0820≈0.01789 SE = \frac{0.08}{\sqrt{20}} \approx 0.01789
step 3
Next, we calculate the z-score for xˉ=5.14 \bar{x} = 5.14 inches using the formula z=xˉ−μSE z = \frac{\bar{x} - \mu}{SE} . Therefore, z=5.14−5.20.01789≈−3.35 z = \frac{5.14 - 5.2}{0.01789} \approx -3.35
step 4
We then look up the z-score of −3.35 -3.35 in the standard normal distribution table, which gives us a probability of approximately 0.00043 0.00043 or 0.043% 0.043\%
step 5
Thus, the percentage of sample means that will be less than 5.14 5.14 inches is 0.043% 0.043\%
A
Key Concept
Central Limit Theorem
Explanation
The Central Limit Theorem states that the distribution of sample means will approach a normal distribution as the sample size increases, allowing us to use z-scores to find probabilities.
Solution by Steps
step 2
According to the Central Limit Theorem, the sampling distribution of the sample mean xˉ \bar{x} will be normally distributed with mean μxˉ=μ=0.9 \mu_{\bar{x}} = \mu = 0.9 seconds and standard deviation σxˉ=σn=0.336=0.05 \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{0.3}{\sqrt{36}} = 0.05 seconds
step 3
We need to find the probability that the sample mean is less than 0.84 seconds, which can be expressed as P(\bar{x} < 0.84) . We first calculate the z-score: z=xˉ−μxˉσxˉ=0.84−0.90.05=−1.2 z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{0.84 - 0.9}{0.05} = -1.2
step 4
Using the z-score, we can find the corresponding probability from the standard normal distribution table: P(Z < -1.2) \approx 0.1151
step 5
Therefore, the probability that the sample mean will be less than 0.84 seconds is approximately 0.1151
B
Key Concept
Central Limit Theorem
Explanation
The Central Limit Theorem states that the distribution of the sample mean approaches a normal distribution as the sample size increases, regardless of the population's distribution. This allows us to use z-scores to find probabilities related to sample means.
Solution by Steps
step 2
The standard deviation of the sampling distribution of the sample mean is calculated using the formula σxˉ=σn \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} . Thus, σxˉ=0.39=0.33=0.1 \sigma_{\bar{x}} = \frac{0.3}{\sqrt{9}} = \frac{0.3}{3} = 0.1
step 3
We need to find the probability that the average length of the steel sheets is more than 29.95 inches, which can be expressed as P(\bar{x} > 29.95) . We first convert this to a z-score using the formula z=xˉ−μσxˉ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} . Substituting the values, we get z=29.95−30.050.1=−0.10.1=−1 z = \frac{29.95 - 30.05}{0.1} = \frac{-0.1}{0.1} = -1
step 4
We now look up the z-score of -1 in the standard normal distribution table, which gives us P(Z < -1) \approx 0.1587 . Therefore, P(Z > -1) = 1 - P(Z < -1) = 1 - 0.1587 = 0.8413
step 5
Thus, the probability that the average length of a steel sheet from a sample of 9 units is more than 29.95 inches long is approximately 0.8413
B
Key Concept
Normal Distribution and Sampling Distribution
Explanation
The Central Limit Theorem states that the sampling distribution of the sample mean approaches a normal distribution as the sample size increases, allowing us to use z-scores to find probabilities.
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