(a) To find the percentage of people with serum alpha tocopherol levels between 400 and 1000 μg/dL, we need to calculate the probability of this range under the given normal distribution.
Z1 = (400 - 860) / 340 = -1.647
Z2 = (1000 - 860) / 340 = 0.412
Using the z-table or a statistical calculator, we can find the cumulative probabilities associated with these z-scores. P(400 < x < 1000) = P(-1.647 < Z < 0.412)
P(-1.647 < Z < 0.412) = P(Z < 0.412) - P(Z < -1.647)
Using a z-table, P(Z < 0.412) = 0.6587 and P(Z < -1.647) = 0.0499.
Therefore, P(400 < x < 1000) = 0.6587 - 0.0499 = 0.6088 or 60.88%.
(b) To find the percentage of people with toxic levels of alpha tocopherol (> 2000 μg/dL), we need to calculate the probability of the serum levels being greater than 2000 μg/dL.
Z = (2000 - 860) / 340 = 3.058
P(X > 2000) = P(Z > 3.058)
Using a z-table, P(Z > 3.058) is approximately 0.0012 or 0.12%.
Therefore, approximately 0.12% of people will be identified as having toxic levels of alpha tocopherol.
(c) To determine if the number of people (4) with toxic levels of serum alpha tocopherol (> 2000 μg/dL) in the study of 2000 people is unusual, we can calculate the probability of observing at least 4 individuals with toxic levels under the assumed normal distribution.
Using the binomial distribution, we can calculate:
P(X ≥ 4) = 1 - P(X ≤ 3)
This can be calculated using a binomial probability calculator or software. Assuming a probability of success (having toxic levels) of 0.0012 and a sample size of 2000, P(X ≥ 4) is very low.
Therefore, the probability of observing 4 or more people with toxic levels is very low, suggesting that it is an unusual number.
(d) To determine if the mean tocopherol levels in the study of 36 people (mean = 1000 μg/dL) is greater than the population mean (μ = 860 μg/dL), we can conduct a hypothesis test.
Null hypothesis (H0): μ = 860
Alternative hypothesis (Ha): μ > 860
Using a t-test, we can calculate the t-value:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (1000 - 860) / (340 / sqrt(36))
t = 140 / (340 / 6)
t = 140 / 56.67
t ≈ 2.47
At a 5% significance level (α = 0.05), the critical t-value for a one-tailed test with 35 degrees of freedom is approximately 1.6909 (from a t-table).
Since the calculated t-value (2.47) is greater than the critical t-value (1.6909), we can reject the null hypothesis.
At a 1% significance level (α = 0.01), the critical t-value for a one-tailed test with 35 degrees of freedom is approximately 2.7087 (from a t-table).
Since the calculated t-value (2.47) is less than the critical t-value (2.7087), we fail to reject the null hypothesis.
Therefore, the mean tocopherol levels in the study of 36 people was not significantly greater than the population mean at the 5% significance level, but it was significantly greater at the 1% significance level.
The significance level that returned the same decision in both cases was 5%.
The significance level that returned a different decision was 1%.
To construct a 99% confidence interval for the population based on the sample mean, we can use the formula:
Confidence interval = sample mean ± (t-value * (sample standard deviation / sqrt(sample size)))
Using a t-table or statistical software, the critical t-value for a 99% confidence interval with 35 degrees of freedom is approximately 2.7307.
Confidence interval = 1000 ± (2.7307 * (340 / sqrt(36)))
Confidence interval = 1000 ± (2.7307 * (340 / 6))
Confidence interval = 1000 ± (2.7307 * 56.67)
Confidence interval ≈ 1000 ± 154.60
The confidence interval is approximately (845.40, 1154.60).
To determine if the result agrees with the 1% significance level for the two-tailed test, we need to check if the population mean (μ = 860) lies within the confidence interval.
Since 860 is within the confidence interval (845.4 to 1154.6), the result agrees with the 1% significance level for the two-tailed test.
A confidence interval cannot be used in a one-tailed test because confidence intervals are two-sided and capture a range of values that are consistent with the data. In a one-tailed test, we are interested in determining if the data is significantly greater than or less than the hypothesized value, and a confidence interval does not provide this specific information. A one-tailed test requires specifying a direction and using critical values of the test statistic for that specific direction.
1.The distribution of serum levels of alpha tocopherol (serum vitamin E) is here assumed to be distributed normal with mean μ=860 μg/dL and standard deviation σ=340 μg/dL.
(a).What percent of people have serum alpha tocopherol levels between 400 and 1000 μg/dL?
(b).Suppose a person is identified has having toxic levels of alpha tocopherol if his or her serum level is > 2000 μg/dL. What percentage of people will be so identified?
(c).A study is undertaken for evidence of toxicity among 2000 people who regularly take vitamin-E supplements. The investigators found that 4 people have serum alpha tocopherol levels > 2000 μg/dL. Is this an unusual number of people with toxic levels of serum alpha tocopherol?
(d).A study is undertaken for evidence of toxicity among 36 people who regularly use vitamin E supplements. The investigators found that the average serum alpha tocopherol levels was equal to 1000 μg/dL. Was the mean tocopherol levels greater than the population mean? (Test at 5% significance level then at 1% significance level).
Was the mean tocopherol level greater than 1000 μg/dL. Again test at 5% significance level and % significance level.
Which significance level returned the same decision in both cases
Which significance level returned a different decision
Construct 99% confidence interval for population based on the sample mean.
Does the result agree to the 1% significance level for the two tailed test
Why can’t you use confidence interval in one tailed test?
5 answers
1.Suppose that, among 25-34 year old males in the general population, the average daily intake of linoleic acid is 15 g. As part of a dietary-instruction program, ten 25-34 year old males adopted a vegetarian diet for one month. While on the diet, the average daily intake of linoleic acid in this sample was a sample mean =13 g with a sample standard deviation = 4 g. Suppose we are uncertain what effect a vegetarian diet will have on the level of linoleic-acid.
(a).What are the null and alternative hypotheses in this case?
(b).Carry out the appropriate statistical hypothesis test to compare the mean level of linoleic acid in the vegetarian population with that of the general population.
(c).Using the summary statistics provided at the beginning of this question (sample mean =13 g and sample standard deviation = 4 g), compute a 90% confidence interval estimate of the true mean intake of linoleic acid in the vegetarian population of 25-34 year old males.
(d).Suppose that the weight W of individual male patients registered at a certain diet clinic is distributed normal with mean μ = 190 and variance σ2 = 100. Consider defined as the sample mean of 25 individual observations of x ̅. Find the values “a” and “b” such that
P(a≤x ̅≤b)=0.80 where x ̅-a=b-x ̅.
(e).You have been hired as the quality control office of a pharmaceutical company that manufactures aspirin tablets. According to the quality assurance protocol, your job is to do the following. For each shipment of possibly thousands and thousands of aspirin tablets, randomly select and test 300 tablets, then accept the batch if there is no more than 13 tablets that don’t meet the test specification. If, at the pharmaceutical company, the probability that randomly selected tablet is defective is .04, what is the probability that a randomly selected shipment of 300 tablets will be accepted?
What is the probability that a randomly selected sample contains 20 that do not meet the requirements?
Is the value 20 unusual (from a above)
(a).What are the null and alternative hypotheses in this case?
(b).Carry out the appropriate statistical hypothesis test to compare the mean level of linoleic acid in the vegetarian population with that of the general population.
(c).Using the summary statistics provided at the beginning of this question (sample mean =13 g and sample standard deviation = 4 g), compute a 90% confidence interval estimate of the true mean intake of linoleic acid in the vegetarian population of 25-34 year old males.
(d).Suppose that the weight W of individual male patients registered at a certain diet clinic is distributed normal with mean μ = 190 and variance σ2 = 100. Consider defined as the sample mean of 25 individual observations of x ̅. Find the values “a” and “b” such that
P(a≤x ̅≤b)=0.80 where x ̅-a=b-x ̅.
(e).You have been hired as the quality control office of a pharmaceutical company that manufactures aspirin tablets. According to the quality assurance protocol, your job is to do the following. For each shipment of possibly thousands and thousands of aspirin tablets, randomly select and test 300 tablets, then accept the batch if there is no more than 13 tablets that don’t meet the test specification. If, at the pharmaceutical company, the probability that randomly selected tablet is defective is .04, what is the probability that a randomly selected shipment of 300 tablets will be accepted?
What is the probability that a randomly selected sample contains 20 that do not meet the requirements?
Is the value 20 unusual (from a above)
(a) The null hypothesis (H0) in this case would be that there is no difference in the mean level of linoleic acid in the vegetarian population compared to the general population. The alternative hypothesis (Ha) would be that there is a difference in the mean level of linoleic acid in the vegetarian population compared to the general population.
(b) To carry out the appropriate statistical hypothesis test, we can use a t-test since the population standard deviation is unknown and we have a small sample size (n < 30).
The null hypothesis (H0) would be: μ = 15 (no difference in the mean level of linoleic acid)
The alternative hypothesis (Ha) would be: μ ≠ 15 (a difference in the mean level of linoleic acid)
Using the sample mean (x̄ = 13 g), the sample standard deviation (s = 4 g), and the population mean (μ = 15 g), we can calculate the t-value:
t = (x̄ - μ) / (s / sqrt(n))
t = (13 - 15) / (4 / sqrt(10))
t = -2 / (4 / sqrt(10))
t ≈ -0.894
Using a t-table or statistical software, we can find the critical t-value for a two-tailed test at a 95% confidence level with 9 degrees of freedom. The critical t-value is approximately 2.262.
Since the calculated t-value (-0.894) is less than the critical t-value (2.262), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the mean level of linoleic acid in the vegetarian population is different from the general population.
(c) To compute a 90% confidence interval estimate of the true mean intake of linoleic acid in the vegetarian population, we can use the formula:
Confidence interval = sample mean ± (t-value * (sample standard deviation / sqrt(sample size)))
Using a t-table or statistical software, the critical t-value for a 90% confidence interval with 9 degrees of freedom is approximately 1.833.
Confidence interval = 13 ± (1.833 * (4 / sqrt(10)))
Confidence interval = 13 ± (1.833 * 1.266)
Confidence interval ≈ 13 ± 2.321
The confidence interval is approximately (10.68, 15.32).
(d) To find the values "a" and "b" such that P(a ≤ x̅ ≤ b) = 0.80, where x̅ - a = b - x̅, we can use the concept of the sampling distribution of the sample mean.
From the given information, we know that the mean (μ) = 190 and the variance (σ^2) = 100.
The standard deviation (σ) = sqrt(σ^2) = sqrt(100) = 10.
To find the z-values corresponding to the desired probability, we need to find the z-value associated with a cumulative probability of 0.40 (0.80 / 2) on each tail.
Using a standard normal distribution table or a z-table, the z-value associated with a cumulative probability of 0.40 is approximately -0.253.
Since x̅ - a = b - x̅, we can rewrite this as b = 2x̅ - a.
To find the values of "a" and "b", we can substitute the z-value and the given mean into the formula:
-0.253 = (x̅ - μ) / (σ / sqrt(n))
-0.253 = (x̅ - 190) / (10 / sqrt(25))
-0.253 = (x̅ - 190) / 2
Simplifying the equation:
(x̅ - 190) / 2 = -0.253
x̅ - 190 = -0.506
x̅ = 189.494
Substituting x̅ back into the formula:
b = 2(189.494) - a
b = 378.988 - a
Therefore, the values "a" and "b" are approximately (189.494, 378.988).
(e) To find the probability that a randomly selected shipment of 300 tablets will be accepted, we can use the binomial probability formula:
P(X ≤ 13) = Σ (nCx) * p^x * (1-p)^(n-x) where n = 300, p = 0.04, and x ranges from 0 to 13.
Using a binomial probability calculator or software, we can calculate P(X ≤ 13) to find the probability that no more than 13 tablets do not meet the requirements.
The probability that a randomly selected sample contains exactly 20 tablets that do not meet the requirements can be calculated using the same binomial probability formula:
P(X = 20) = (nC20) * p^20 * (1-p)^(n-20) where n = 300 and p = 0.04.
To determine if the value of 20 is unusual, we can compare it to the probability of observing 20 or more tablets that do not meet the requirements:
P(X ≥ 20) = 1 - P(X ≤ 19)
Using a binomial probability calculator or software, we can calculate P(X ≥ 20) to find the probability of observing 20 or more tablets that do not meet the requirements. If this probability is very low, we can conclude that the value of 20 is unusual.
(b) To carry out the appropriate statistical hypothesis test, we can use a t-test since the population standard deviation is unknown and we have a small sample size (n < 30).
The null hypothesis (H0) would be: μ = 15 (no difference in the mean level of linoleic acid)
The alternative hypothesis (Ha) would be: μ ≠ 15 (a difference in the mean level of linoleic acid)
Using the sample mean (x̄ = 13 g), the sample standard deviation (s = 4 g), and the population mean (μ = 15 g), we can calculate the t-value:
t = (x̄ - μ) / (s / sqrt(n))
t = (13 - 15) / (4 / sqrt(10))
t = -2 / (4 / sqrt(10))
t ≈ -0.894
Using a t-table or statistical software, we can find the critical t-value for a two-tailed test at a 95% confidence level with 9 degrees of freedom. The critical t-value is approximately 2.262.
Since the calculated t-value (-0.894) is less than the critical t-value (2.262), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the mean level of linoleic acid in the vegetarian population is different from the general population.
(c) To compute a 90% confidence interval estimate of the true mean intake of linoleic acid in the vegetarian population, we can use the formula:
Confidence interval = sample mean ± (t-value * (sample standard deviation / sqrt(sample size)))
Using a t-table or statistical software, the critical t-value for a 90% confidence interval with 9 degrees of freedom is approximately 1.833.
Confidence interval = 13 ± (1.833 * (4 / sqrt(10)))
Confidence interval = 13 ± (1.833 * 1.266)
Confidence interval ≈ 13 ± 2.321
The confidence interval is approximately (10.68, 15.32).
(d) To find the values "a" and "b" such that P(a ≤ x̅ ≤ b) = 0.80, where x̅ - a = b - x̅, we can use the concept of the sampling distribution of the sample mean.
From the given information, we know that the mean (μ) = 190 and the variance (σ^2) = 100.
The standard deviation (σ) = sqrt(σ^2) = sqrt(100) = 10.
To find the z-values corresponding to the desired probability, we need to find the z-value associated with a cumulative probability of 0.40 (0.80 / 2) on each tail.
Using a standard normal distribution table or a z-table, the z-value associated with a cumulative probability of 0.40 is approximately -0.253.
Since x̅ - a = b - x̅, we can rewrite this as b = 2x̅ - a.
To find the values of "a" and "b", we can substitute the z-value and the given mean into the formula:
-0.253 = (x̅ - μ) / (σ / sqrt(n))
-0.253 = (x̅ - 190) / (10 / sqrt(25))
-0.253 = (x̅ - 190) / 2
Simplifying the equation:
(x̅ - 190) / 2 = -0.253
x̅ - 190 = -0.506
x̅ = 189.494
Substituting x̅ back into the formula:
b = 2(189.494) - a
b = 378.988 - a
Therefore, the values "a" and "b" are approximately (189.494, 378.988).
(e) To find the probability that a randomly selected shipment of 300 tablets will be accepted, we can use the binomial probability formula:
P(X ≤ 13) = Σ (nCx) * p^x * (1-p)^(n-x) where n = 300, p = 0.04, and x ranges from 0 to 13.
Using a binomial probability calculator or software, we can calculate P(X ≤ 13) to find the probability that no more than 13 tablets do not meet the requirements.
The probability that a randomly selected sample contains exactly 20 tablets that do not meet the requirements can be calculated using the same binomial probability formula:
P(X = 20) = (nC20) * p^20 * (1-p)^(n-20) where n = 300 and p = 0.04.
To determine if the value of 20 is unusual, we can compare it to the probability of observing 20 or more tablets that do not meet the requirements:
P(X ≥ 20) = 1 - P(X ≤ 19)
Using a binomial probability calculator or software, we can calculate P(X ≥ 20) to find the probability of observing 20 or more tablets that do not meet the requirements. If this probability is very low, we can conclude that the value of 20 is unusual.
1.Suppose that, among 25-34 year old males in the general population, the average daily intake of linoleic acid is 15 g. As part of a dietary-instruction program, ten 25-34 year old males adopted a vegetarian diet for one month. While on the diet, the average daily intake of linoleic acid in this sample was a sample mean =13 g with a sample standard deviation = 4 g. Suppose we are uncertain what effect a vegetarian diet will have on the level of linoleic-acid.Suppose that the weight W of individual male patients registered at a certain diet clinic is distributed normal with mean μ = 190 and variance σ2 = 100. Consider defined as the sample mean of 25 individual observations of x ̅. Find the values “a” and “b” such that
P(a≤x ̅≤b)=0.80 where x ̅-a=b-x ̅.
P(a≤x ̅≤b)=0.80 where x ̅-a=b-x ̅.
To find the values "a" and "b" such that P(a ≤ x̅ ≤ b) = 0.80, where x̅ - a = b - x̅, we can use the concept of the sampling distribution of the sample mean.
From the given information, we know that the mean (μ) = 190 and the variance (σ^2) = 100.
The standard deviation (σ) = sqrt(σ^2) = sqrt(100) = 10.
To find the z-values corresponding to the desired probability, we need to find the z-value associated with a cumulative probability of 0.10 on each tail.
Using a standard normal distribution table or a z-table, the z-value associated with a cumulative probability of 0.10 is approximately -1.282.
Since x̅ - a = b - x̅, we can rewrite this as b = 2x̅ - a.
To find the values of "a" and "b", we can substitute the z-value and the given mean into the formula:
-1.282 = (x̅ - μ) / (σ / sqrt(n))
-1.282 = (x̅ - 190) / (10 / sqrt(25))
-1.282 = (x̅ - 190) / 2
Simplifying the equation:
(x̅ - 190) / 2 = -1.282
x̅ - 190 = -2.564
x̅ = 187.436
Substituting x̅ back into the formula:
b = 2(187.436) - a
b = 374.872 - a
Therefore, the values "a" and "b" are approximately (187.436, 374.872).
From the given information, we know that the mean (μ) = 190 and the variance (σ^2) = 100.
The standard deviation (σ) = sqrt(σ^2) = sqrt(100) = 10.
To find the z-values corresponding to the desired probability, we need to find the z-value associated with a cumulative probability of 0.10 on each tail.
Using a standard normal distribution table or a z-table, the z-value associated with a cumulative probability of 0.10 is approximately -1.282.
Since x̅ - a = b - x̅, we can rewrite this as b = 2x̅ - a.
To find the values of "a" and "b", we can substitute the z-value and the given mean into the formula:
-1.282 = (x̅ - μ) / (σ / sqrt(n))
-1.282 = (x̅ - 190) / (10 / sqrt(25))
-1.282 = (x̅ - 190) / 2
Simplifying the equation:
(x̅ - 190) / 2 = -1.282
x̅ - 190 = -2.564
x̅ = 187.436
Substituting x̅ back into the formula:
b = 2(187.436) - a
b = 374.872 - a
Therefore, the values "a" and "b" are approximately (187.436, 374.872).
1 answer