One classical experiment in extrasensory perception tests the ability of an individual to show telepathy - to read the mind of another individual. This test uses five cards with different designs, all known to both participants. In a trial, the 'sender' sees a randomly chosen card and concentrates on the design. The 'receiver' attempts to guess the identity of the card. Each of the five cards is equally likely to be chosen, and only one card is the correct answer at any point
Which of the following is an appropriate null hypothesis for this telepathy experiment?
The proportion of correctly identified cards by the receiver is greater than 0.2 |
||
The proportion of correctly identified cards by the receiver is 0.2 |
||
The proportion of correctly identified cards by the receiver is something other than 0.2 |
||
The proportion of correctly identified cards by the receiver is less than 0.2 |
Which of the following is an appropriate alternative hypothesis for this telepathy experiment, if a two tailed test is to be used?
The proportion of correctly identified cards by the receiver is greater than 0.2 |
||
The proportion of correctly identified cards by the receiver is 0.2 |
||
The proportion of correctly identified cards by the receiver is something other than 0.2 |
||
The proportion of correctly identified cards by the receiver is less than 0.2 |
Which of the following is an appropriate alternative hypothesis for this telepathy experiment, if a one tailed test is to be used?
The proportion of correctly identified cards by the receiver is greater than 0.2 |
||
The proportion of correctly identified cards by the receiver is 0.2 |
||
The proportion of correctly identified cards by the receiver is something other than 0.2 |
||
The proportion of correctly identified cards by the receiver is less than 0.2 |
In this experiment we are really trying to see if mind reading occurs. That is we are trying to prove that the sender can actually tell telepathically the receiver what card he is seeing. True or false: In this context we should really use a one tailed test.
True
False
Out of 10 trials, a receiver got four cards correct. What is the probability of that particular outcome (4 correct guesses out of 10) if the receiver was guessing, that is picking the cards at random? Round off to THREE digits after the decimal point.
Out of 10 trials, a receiver got four cards correct. What is the probability of that particular outcome (4 correct guesses out of 10) if the receiver was guessing, that is picking the cards at random? Round off to THREE digits after the decimal point.
Out of 10 trials, a receiver got four cards correct. (This is identical to question 5). What is the P-value from a two tailed binomial test corresponding to this result? Round off to TWO digits after the decimal point. (Write down the answer, you will need it for the next questions.
8) Out of 10 trials, a receiver got four cards correct. (This is identical to question 5). What is the lower bound of the Agresti-Coull confidence interval for the proportion of correctly identified cards? Round off to TWO digits after the decimal point.
9) Out of 10 trials, a receiver got four cards correct. (This is identical to question 5). What is the upper bound of the Agresti-Coull confidence interval for the proportion of correctly identified cards? Round off to TWO digits after the decimal point.
10) Based on the result from the one tailed binomial test do you reject the null hypothesis?
11) Based on the result of the one tailed binomial test, which of the following is an appropriate conclusion from the observation of 4 correctly identified cards out of 10 trials?
One classical experiment in extrasensory perception tests the ability of an individual to show telepathy -...
prob 24 One test for claims of extrasensory perception (ESP) involves using Zener cards. Each card shows one of five different symbols (square, circle, star, cross, wavy lines), and the person being tested has to predict the shape on each card before it is selected. Using the Distribution tool below, find each of the probabilities requested for a person who has no ESP and is just guessing. Standard Normal Distribution Mean 0.0 Standard Deviation 1.0 5000 5000 0.0 0,0000 1.0...
Trials in an experiment with a polygraph include 98 results that include 24 cases of wrong results and 74 cases of correct results. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80% of the time. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the nuill hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial...
Trials in an experiment with a polygraph include 97 results that include 22 cases of wrong results and 75 cases of correct results. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80% of the time. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial...
2. (2 True-False. Just say whether each statement is True or False – no need to justify your answer. 1. If the number of trials in the binomial distribution increases by 1 (and P equals .50), the probability of getting either of the most extreme possible outcomes (that is, 0 or N) is cut in half. 2. If the number of trials in the binomial distribution increases by 1 (and P does not equal .50), the probability of getting either of...
Question 6 (1 point) In JMP, to conduct a Hypothesis Test about one proportion using a Z-test Use Analyze > Distribution. Then select the variable and then at the red triangle, select Test Probabilities and enter the hypothesized value in the dialogue box. Go to Add Ins. Select Hypothesis Test for One Proportion from the list, select Raw or Summarized data, pick the column that contains the variable and then type in the value associated with success EXACTLY the way...
In a program designed to help patients stop smoking, 236 patients were given sustained care, and 84.3% of them were no longer smoking after one month. Use a 0.05 significance level to test the claim that 80% of patients stop smoking when given sustained care. Identify the null and alternative hypotheses for this test. Choose the correct answer below. O A. Ho:p*0.8 H: p=0.8 OB. Ho: P = 0.8 H:p<0.8 @ C. Ho: p = 0.8 H:p*0.8 OD. Ho: p=0.8...
[15 marks] Suppose that students enrolled in one course are required to take four tests, and each student’s final grade for this course is the average of his/her grades of these four tests. This question asks you to write a program that can be used to compute the lowest final grade, highest final grade and the average final grade. General Requirements: Use a5q1.c as the name of your C source code file. We will use the following command on bluenose...
plesse help with explanations and answers for all of those ... im stuck and cant figure out how to do them. its the hard copy from web work wnd they are all incorrect. P align="center Inference about a Population Propor- tion /p hr Due: 07/01/2019 at 11:59pm EDT hrStudents will be able to: iUL ili;, Perform a hypothe- sis test on population proportion /li ili; Cakculate a confidence interval for a population proportionAi Interpret levels of significance/i il Perform a...
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Provide an appropriate response. 1) Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority of Americans. A random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a 98% confidence interval. Express the...
Please m asking for assistance on statistics Q10. A population consists of the following N-5 scores: 0, 6, 4, 3, and 12 a. Compute u ando for the population b. Find the 2-score for each score in the population c. Transform the original population into a new population of N = 5 scores with a mean of u = 100 and a standard deviation of a = 20 Q11. For a sample with a mean of M = 51, a...