1.3 HW Stats

Suppose a friend of yours says she is a 75% free-throw shooter in basketball. You don't think she is that good and want to test her to gather evidence that she makes less than 75% of her free throws in the long run. You have her shoot 40 free throws and she makes 26 (or 65%) of them. Based on this information, you make the following null distribution. What is the value of the standardized statistic for your friend?

-1.47

c. Calculate and report the value of the standardized statistic. (Round answer to 2 decimal places, e.g. 1.75)

-.581

c. Use the One Proportion applet to generate the null distribution of the proportion of "successes." Which of the following is closest to the value of the standard deviation of your null distribution?

0.14

Although we know some animals, like dogs, have a keen sense of smell, what about humans? A woman from Scotland, Joy Milne, claimed that she was able to determine whether or not someone had Parkinson's disease just based on their smell (Morgan, J., 2016). Her husband died from the disease and she noticed that he started smelling different prior to his diagnosis. To test this, researchers had six people known to have Parkinson's disease and six people thought not to have the disease wear t-shirts for a day. They collected the shirts and then tested Milne. She correctly identified 11 of the 12 shirts. a. Let π represent the probability that Joy Milne can correctly determine whether someone has Parkinson's disease from smelling their shirt. What is the value of π if she is just guessing?

0.50

Consider the output given below that was obtained using the One Proportion applet. Use information from the output to find the standardized statistic for a sample proportion value of 0.45. (Round answer to two decimal places, e.g. 1.23) prob of success: 0.30 sample size (n): 25 number of samples: 1000

1.64

How many standard deviations above the mean of the null distribution is the sample proportion? (i.e., What is the standardized statistic for this test?). Round answer to 2 decimal places, e.g. 2.23.

1.72

d. Determine the standardized statistic for the observed sample proportion of "successes." (Round answer to 2 decimal places, e.g. 1.75)

2.94

c. Calculate and report the value of the standardized statistic. (Round answer to 2 decimal places, e.g. 1.75)

5.80

f. Based on the statistic from this test (67/141 = 0.475) how can you determine that you will get a very large p-value without even developing a null distribution? Choose the best among the following statements. Multiple choice 3 Question 8

Because 0.475 is less than 0.50 and in the opposite direction than that conjectured by the alternative hypothesis.

e. Based on the standardized statistic, what conclusion you would draw about the research conjecture that Milne has the ability to detect Parkinson's disease at a rate higher than 50%? Choose the best among the following statements.

Because the standardized statistic is larger than 2, we have evidence to claim that Joy Milne's probability of correctly determining whether someone has Parkinson's disease is greater than 0.50.

Blindsight is a condition in which people are blind but can still respond to things they cannot consciously see. A patient (who had suffered brain damage and whom researchers Persaud et al. called GY in Nature Neuroscience) was right-side blind, meaning he could not see anything in his right field of vision. But could GY still "see" objects in his right field of vision even though he has no conscious perception of the objects? Researchers tested GY on this by having him face a video monitor where they would run trials by sometimes having a square appear on the right side of the monitor and sometimes not. In 200 trials, GY correctly determined whether the square appeared or not 141 times. To determine if GY is more likely than random chance to correctly identify whether or not the square was present answer the following. a. Set up the correct null and alternative hypotheses in symbols.

H0: π = 0.50 Ha: π >0.50

Recall the previous question about GY and blindsight. To test GY's awareness of whether he correctly identified whether the square was present or not, they had him wager on whether or not he thought he was correct after he made his choice. He had to bet either a British pound or half a pound on whether his guess was correct. If he was correct, that amount was added to his earnings, and if he was not correct, that amount was subtracted. So, he should make the high wager when he was more confident in his choice. In the 141 times he correctly identified whether the square was present or not, he made the high wager 67 times. When he has made a correct choice, let's find out if GY is more likely to make the high wager than the small wager (and thus have some real conscious awareness of the object being present or not) by answering the following. a. Set up the correct null and alternative hypotheses in symbols.

H0: π = 0.50 Ha: π >0.50

b. Set up the correct null and alternative hypotheses in symbols to test this scenario.

H0: π = 0.50 Ha: π > 0.50

f. Milne was adamant that one of the shirts came from someone with the disease when, in fact, that person had not been diagnosed. However, eight months later, the person she claimed had Parkinson's was given a diagnosis that he had the disease. So she, in fact, got all 12 correct! Using this new statistic of 12 out of 12, determine the standardized statistic for the observed sample proportion of "successes." Does this now give stronger or weaker evidence that she is doing better than just guessing than if she only got 11 out of 12 correct like what was originally thought?

Stronger evidence, because the standardized statistic is around 3.5 which is farther from 0 than the previous standardized statistic.

Suppose that a standardized statistic (standardized sample proportion) for a study is calculated to be 2.45. Which of the following is the most appropriate interpretation of this standardized statistic?

The observed value of the sample proportion is 2.45 SDs above the hypothesized parameter value.

e. Using the applet, find a p-value. Does the p-value lead you to the same conclusion as the standardized statistic did? Choose the best among the following statements.

The p-value is approximately 0.75 and it leads to the same conclusion as the one from the standardized statistic.

e. Using the applet, find a p-value. Does the p-value lead you to the same conclusion as the standardized statistic did? Choose the best among the following statements.

The p-value is very small (< 0.001) and it leads to the same conclusion as the one from the standardized statistic.

d. Based on the standardized statistic is there strong evidence that GY is more likely than random chance to make a high wager when he makes a correct identification? Choose the best of the following statements.

There is not strong evidence that GY is more likely than random chance to make a high wager when he makes a correct identification because the standardized statistic is close to zero.

d. Based on the standardized statistic is there strong evidence that GY is more likely than random chance to correctly identify whether or not the square was present (or that GY has blindsight)? Choose the best of the following statements.

There is very strong evidence that GY is more likely than random chance to correctly identify whether or not the square was present because the standardized statistic is greater than 3.

Suppose you are testing the hypotheses H0: π = 0.50 and Ha: π > 0.50 and get a sample proportion of 0.65. From this, you compute a standardized statistic and a p-value. If your sample proportion had been 0.70 then your standardized statistic would Answer 1 Question 4 decreaseincrease and your p-value would Answer 2 Question 4

increase, decrease

Based on your p-value, do you have strong evidence that students prefer Chips Ahoy over Chipsters?

no, greater, 0.05

Based on your standardized statistic, do you have strong evidence that students prefer Chips Ahoy over Chipsters? (Fill in the following blanks to answer this.)

no, less, 2

Student researchers were interested in whether people will be more likely to choose the name-brand cookie (Chips Ahoy) over the store-brand (Chipsters) in a blind taste test. They tested this with 20 subjects and found that 14 (70%) chose Chips Ahoy as their favorite. They conducted a test of significance using simulation and got the following null distribution. (Note that this null distribution uses only 100 simulated samples and not the usual 1000 or 10,000.) Let π represent the long-run proportion of students that will pick Chips Ahoy as their favorite. a) Using the null distribution above, what is the p-value when testing H0: π = 0.50 versus an alternative of Ha: π > 0.50.

p-value = 6/100 = 0.06

b. Using the One Proportion applet, construct an appropriate null distribution. What statistic is represented in the null distribution, and what is the standard deviation of your null distribution? The statistic is the Answer 3 Question 7 standardized statisticsample meanp-valuesample proportion and the standard deviation of the null distribution is Answer 4 Question 7 0.0440.0350.016

sample proportion, 0.035

b. Using the One Proportion applet, construct an appropriate null distribution. What statistic is represented in the null distribution, and what is the standard deviation of this null distribution? The statistic is the Answer 3 Question 8 sample meanstandardized statisticsample proportionp-value and the standard deviation of the null distribution is Answer 4 Question 8 0.0150.0430.061.

sample proportion, 0.043