B. the standard deviation of the sampling distribution for the statistic. C. the total sum of squares of deviations of the observations about the mean. D. the number of standard deviations that a statistic value differs from the parameter value. 8. A theoretical sampling distribution of a statistic consists of A. the results of a sample.
A. 79% and 500 days B. 13% and 79% C. 79% and one in twelve D. 13% and one in twelve ------------------------------------------------- A sufficiently large coverage error will result in which of the following? A. Statistics about the actual population rather than the target population B. Non-response bias C. Inability to perform inferential statistics D. Probability sampling When every member of a population has the chance of being selected based on the probability, or frequency, of its representation in that population, you are using which type of sampling?
* Question 1 2 out of 2 points | | | Probability trees are used only to compute conditional probabilities. Answer | | | | | Selected Answer: | False | Correct Answer: | False | | | | | * Question 2 2 out of 2 points | | | Seventy two percent of all observations fall within 1 standard deviation of the mean if the data is normally distributed. Answer | | | | | Selected Answer: | False | Correct Answer: | False | | | | | * Question 3 2 out of 2 points | | | If two events are not mutually exclusive, then P(A or B) = P(A) + P(B) Answer | | | | | Selected Answer: | False | Correct Answer: | False | | | | | * Question 4 2 out of 2 points | | | The maximin approach involves choosing the alternative with the highest or lowest payoff. Answer | | | | | Selected Answer: | False | Correct Answer: | False | | | | | * Question 5 2 out of 2 points | | | Both maximin and minimin criteria are optimistic. Answer | | | | | Selected Answer: | False | Correct Answer: | False | | | | | * Question 6 2 out of 2 points | | | The Hurwicz criterion is a compromise between the maximax and maximin criteria.
A. True B. False The p-value of .05 indicates that if Ho were true, a sample result as extreme as the one calculated would happen by chance approximately 5 times out of 1000. A. True B.
The first class in a relative frequency table is 50–59 and the corresponding relative frequency is 0.2. What does the 0.2 value indicate? Answer: 0.2 is equal to 1/5 or 20%, 0.2 indicates 20% of the data values are in this particular interval. 3. When you add the values 3, 5, 8, 12, and 20 and then divide by the number of values, the result is 9.6.
1. A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null hypothesis indicates a two-tailed test and the researcher decided to use the 0.10 significance level. For what values of t will the null hypothesis not be rejected?
Explain the difference in these two readings. 8. In the legend beneath Figure 2, the authors give an equation indicating that diastolic blood pressure is DBP = 25.8 + 0.13x. If the value of x is postnatal age of 30 hours, what is the value for Yˆ for neonates ≤ 1,000 grams? Show your calculations.
Hedging Currency Risk at AIFS: Assignment questions: Q. What gives rise to the currency exposure at AIFS? A. AIFS is a company that specializes in providing educational and cultural exchange programs for college and high school students. Overall the company provides services to approximately 50,000 students each year, and has revenues of about $200,000,000. In the assigned case, we study two divisions of AIFS; a College division, a High School Travel Division, and the possible consequences to AIFS’s business exposure with respect to the fluctuations in foreign currency values and other factors.
1. A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null hypothesis indicates a two-tailed test and the researcher decided to use the 0.10 significance level. For what values of t will the null hypothesis not be rejected?
The random sample of 65 satisfaction rating yields a sample mean of x = 42.954. Assuming that S = 2.64, use critical values to test H0 versus Ha at each of a = 10, .05, .01 and .00l. MU = 42, N = 65, X-bar = 42.954, sigma = 2.64 Z= (x-bar – mu)/(sigma/sqrt n) Z = (42.954 – 42)/(2.64/sqrt65 = 2.9134. Wk4/Assignment 9.13 continued Critical upper tail = Z – scores for 1.2816, 1.6449, 2.3263 and 3.092 for a = 0.10, 0.05, 0.01 and 0.001. Since 2.9134>1.2816, 1.6449 and 2.3263, I rejected H0 and accepted Ha at 1 = 0.10, 0.05, and 0.01 and concluded that the mean rating exceeds 42.