When you add the values 3, 5, 8, 12, and 20 and then divide by the number of values, the result is 9.6. Which term best describes this value: average, mean, median, mode, or standard deviation? Answer: 9.6 is the average of the numbers listed and is also the mean of this data. 4. Answer the next four questions using the following set of numbers.
RES/342--- This produced a 28/30 on the final 1) What are the critical z-values for a two-tailed hypothesis test if the significant level = 0.01? C. ± 2.58 2) In classical hypothesis testing, the test statistic is to the critical value what the __________. A. ‘p-value’ is to alpha 3) For a hypothesis test of a single population mean at 95% confidence level, a calculated Z score of 1.7 supports the conclusion that A. the population mean is greater than the hypothesized value 4) If the paired differences are normal in a test of mean differences, then the distribution used for testing is the C. student distribution 5) One hundred women were polled and 60 reported successfully communicating an automobile problem to an auto repairman. A sample of 150 men had 95 reporting the same success.
A divide by 100 B divide by 1000 C multiply by 100 D multiply by 1000 15 Which of the following is the smallest? A 0.07 7 C ----10 1 B -7 D 73% 7 A ----12 7 C ----16 7 B -9 7 D -8 5 Which one of these decimals could be rounded to 34.71? A 34.715 B 34.7 C 34.707 D 34.7039 1 6 What is -- as a percentage? 6 A 16% B 1.6% -C 16 2 % 3 D 60% 7 What is 37.5% of one day? A 10 hours B 8.6 hours C 9.6 hours D 9 hours 8 Increase $38 by 15%.
Suppose we are interested in comparing means from two independent samples. The mean of the first sample is 9 and the mean of the second sample is 17. Let’s assume that the two group means have the same standard errors, equal to 2.5. The 95 percent confidence interval for the first group mean can be calculated as: 9 ± 1.96 × 2.5 where 1.96 is the critical t-value. The confidence interval for the first group mean is thus (4.1, 13.9).
Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10. What proportion of the households spent between $5.00 and $9.00? (Points : 4) .7611 .7628 .0017 .7594 | 5. The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. A sample of four metal sheets is randomly selected from a batch.
Similarly the seventh row of results shows that for a five-year holding period and a “middle level” of risk aversion, the optimal allocation is 40.2% in the risk-free, 41.4% in T-Bonds, 3.2% in World Bonds, 9.0% in Large Stocks and 6.3% in Small Stocks. This portfolio’s expected return is 4.71% and its standard deviation is 2.38%. As the target date approaches the allocation shifts from the higher risk asset classes to the lower risk asset classes to a majority in the risk-free asset. 3. Look at the results in Table 1.
In order to find an approximate result, it can be completed by the function F (x) = x/18 for 0 < x < 6 Therefore, the distribution function is F (x) = x-square divided by 36 for 0 < x < 6 If we set this equal to another random number r1 that is between 0 and then R1 = x-square divided by 36 which results to x = 6√ r1 3. In Excel, use a suitable method for simulating the lost revenue for each day the copier is out of service. Because the amount of copies sold each day is a uniform probability distribution between 2000 to 8000 copies, I designated r3 as a random number between 2000 to 8000. In order to find the amount of business lost on a certain day, I took r3 X (repair time), and the lost revenue
Her combined homework grade was 77. The five regular tests count for 60% of the final grade, the final exam counts for 10%, the project counts for 15%, and homework counts for 15%. What is her weighted mean grade? Test Grades (92+83+77+84+82)/5=83.3 Test grade 83.3x 0.60= 49.98 Final exam 88 x 0.10= 8.8 Class project 95 x 0.15= 14.25 Homework 77 X 0.15= 11.55 Total Grade 84.58 4. Convert Bao’s height to a z-score.
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.
Assign a random variable x, where the value of x is equal to the number of heads observed in the flips. 6) What is the probability of observing 30 or more heads in 50 flips? 7) What is the probability of observing more than 35 tails in 50 flips? Use the following scenario to answer the next 3 questions. Suppose a student in Business Statistics has no been attending class this quarter but decides to take the exam anyway.