Practice Test 4 –Bus 2023 Directions: For each question find the answer that is the best solution provided. There is only one correct answer. 1. For a sample size of 30, changing from using the standard normal distribution to using the t distribution in a hypothesis test, a. will result in the rejection region being smaller b. will result in the rejection region being larger c. would have no effect on the rejection region d. Not enough information is given to answer this question. ANSWER: a -recall that all the t-values are larger than the z-values so it makes it more difficult to reject the null.
The total risk score is 4.14, the greatest relative or standardized difference between pretest and 3 month outcomes. This t ratio has a statistical significance of 0.05 - the least acceptable value for statistical significance. Also the larger the t ratio, the smaller the observed p value and increased odds of being able to reject the null hypothesis. 3. Which t-ratio listed in Table 3 represents the smallest relative difference between the pretest and 3 months?
TOPIC 8 Chi-Square goodness-of-fit test Problem 12.1 Use a chi-square goodness-of-fit to determine whether the observed frequencies are distributed the same as the expected frequencies (α = .05) Category | fo | fe | 1 | 53 | 68 | 2 | 37 | 42 | 3 | 32 | 33 | 4 | 28 | 22 | 5 | 18 | 10 | 6 | 15 | 8 | Step 1 Ho: The observed frequencies are distributed the same as the expected frequencies Ha: The observed frequencies are not distributed the same as the expected frequencies Step 2 df = k – m – 1 Step 3 α = 0.05 x 2 0.05, 5df = 11.0705 Step 4 Reject Ho if x 2 > 11.0705 Category | fo | fe | | 1 | 53 | 68 | | 2 | 37 | 42 | | 3 | 32 | 33 | | 4 | 28 | 22 | | 5 | 18 | 10 | | 6 | 15 | 8
(b) One correct answer to this part is as follows: We can say that we are 95 percent confident the average balance for all overdue bills is in the range from $480 to $520. Since the standard error SE for the sample mean should equal 100/square root of 100 = 10, the margin of error would be twice this value, or $20. This means $480 to $520 is a 95 percent confidence interval for the average of all overdue balances. Another correct answer to this part is as follows: Instead of the average balance, we can find the percentage of all balances that fall in this interval. When we think of all balances, $480 is 0.2 standard deviations below the average (480 -500)/100 = - 0.20), and $520 is 0.2 standard deviations above the average (520 -500)/100 = 0.20).
For what values of t will the null hypothesis not be rejected? a) To the left of -1.645 or to the right of 1.645 b) To the left of -1.345 or to the right of 1.345 c) Between -1.761 and 1.761 d) To the left of -1.282 or to the right of 1.282 2. Which of the following is a characteristic of the F distribution? a) Normally distributed b) Negatively skewed c) Equal to the t-distribution d) Positively skewed Complete Answers here QNT 561 Final Exam 3. For a chi-square test involving a contingency table, suppose the null hypothesis is rejected.
According to the statistics from the traditional method of recruitment, the p-values listed the correlation as very low between the things that were focused on during the interview and recruitment. The interview score with a high p-value and a very low correlation had to be the least relevant predictor to focus on and yet is being used within the traditional method. According to the statistics that were found from the proposed method data education had the best correlation at .14 and a p-value < .01 and although education is seen as statistically significant it is not practically significant. The best predictors were seen to be biodata questionnaire & essay, with a correlation of .22 and a p-value of <.01 for citizenship and a correlation of -.17 and a p-value of <.01, and the personality exam which measures extraversion and conscientiousness focusing on the conscientiousness with a correlation of .18 and p-value of <.01 for citizenship and a correlation of -.33 and a p-value of <.01 for absence. The only issue with the biodata is that it is very expensive at $10 an applicant and due to it being score by a computer lacks the personal quality that Tanglewood is all about.
The (C1)(V1)=(C2)(V2) formula was used to determine how much to dilute the 0.1 g/L stock starch solution. These values produced a standard curve between concentration and their observed absorbance levels in the spectrophotometer. (0.1g/L)(x)=(0g/L)(4mL) X=0mL stock starch *(blank control) 4mL H20 1mL Lugol’s reagent *(must be added for spectrophotometer to read absorbency, not counted in final volume) (0.1g/L)(x)=(0.025g/L)(4mL) X=1mL stock starch 3mL H2O 1mL Lugol’s reagent (0.1g/L)(x)=(0.050g/L)(4mL) X=2mL stock starch 2mL H2O 1mL Lugol’s reagent (0.1g/L)(x)=(0.075g/L)(4mL) X=3mL stock starch 1mL H2O 1mL Lugol’s reagent (0.1g/L)(x)=(0.10g/L)(4mL) X=4mL stock starch 0mL H2O 1mL Lugol’s reagent After creating the known solutions from the concentrations found, the spectrophotometer was zeroed out (see TA for specific instructions on zeroing out spectrophotometer). Each known sample was then placed within the spectrophotometer and the corresponding absorbance levels were recorded and graphed, making a standard curve (see attached graph “Concentration vs. Absorbance”). For each unknown solution, 4mL of each solution was mixed with 1mL of
Then finally I divide all 3 by 703 in order to isolate W. 128.8 < W < 166.7 So people with the height of 73 inches could have a longer than average life span if they weigh between 129 and 167 pounds. The next equation 23<703WH^2<25 I first multiply all 3 terms by H^2 in order to take our the denominator. 23H^2 < 703W < 25H^2 Then I divide all terms by 703 to isolate the W. 23H^2703<W< 25H^2703 After doing this it now becomes an equivalent inequality for me to solve the second weight. I now use my height squared to find the second weight interval. 23(5329)703<W< 25(5329)703 After I multiply I end up with 122567703<W< 133225703 Then I simply divide and end up with 174 < W < 189.5.
Merck will license the drug as long as the expected payoff is greater than the alternative option of not licensing the drug, i.e. 43.3*P1+(-30)*(1-P1)>0 P1>0.41 Therefore, as long as the probability of success in phase 1 is between 0.41 and 1, the decision to license the drug remains the same. Question 4: When the launching cost is $225M, Merck will still choose license the drug. The new decision tree is updated and shown in Exhibit 2. The expected present value of potential cash flow is $10.76 if the drug is licensed.
On-line Science Simulations - Marble Chip Student Worksheet When hydrochloric acid is added to calcium carbonate (marble chips) carbon dioxide is evolved. In the experiment shown, 21.6 g of calcium carbonate is added to 200 cm3 of hydrochloric acid. The number of chips can be varied but the total mass is always 21.6 g and the volume of acid is always 200 cm3 but the concentration can be varied. The apparatus is placed on a balance that has been zeroed so that it always shows the same initial mass at the start of the experiment. During the reaction, carbon dioxide is evolved and the mass decreases.