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(10 points) A new flu vaccine was developed to protect elderly people against viral infection during the winter season. In 120 patients of thepatients who were va...

Question

(10 points) A new flu vaccine was developed to protect elderly people against viral infection during the winter season. In 120 patients of thepatients who were vaccinated. Of the 70 patients, 56 of the patients had no flu symptoms. 50 of the patients who were not vaccinated; 28 patients have no flu symptoms_a) Determine the confidence interval (CI) at 95% (2*=1.96) level for the probability that a patient had no flu symptoms when new vaccine is used during the winter season. What is the sample

(10 points) A new flu vaccine was developed to protect elderly people against viral infection during the winter season. In 120 patients of thepatients who were vaccinated. Of the 70 patients, 56 of the patients had no flu symptoms. 50 of the patients who were not vaccinated; 28 patients have no flu symptoms_ a) Determine the confidence interval (CI) at 95% (2*=1.96) level for the probability that a patient had no flu symptoms when new vaccine is used during the winter season. What is the sample proportion (p) of patients with no symptoms when vaccine is used? CI= b) Carry out a statistical test at the a=0.05 level for whether the probability of vaccinated patients with no flu symptoms higher than the proportion of people who are not vaccinated Determine the null and alternative hypothesis_ What is the pooled sample proportion (p)?



Answers

Business Ethics: Privacy Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that computers and the Internet are reducing privacy. A survey conducted by Peter D. Hart Research Associates for the Shell Poll was reported in USA Today. According to the survey, $37 \%$ of adults are concerned that employers are monitoring phone calls. Use the binomial distribution formula to calculate the probability that (a) out of five adults, none is concerned that employers are monitoring phone calls. (b) out of five adults, all are concerned that employers are monitoring phone calls. (c) out of five adults, exactly three are concerned that employers are monitoring phone calls.

Right. We have a sample of size and equals 209. And in that sample we see 16 objects or sample numbers which satisfy a certain observational criteria. And we want to see that is r equals 16. We want to use this data to test the claim that the population proportion P is less than 0.12 at a confidence level of 1% alpha equals 10.1 Now that we've identified the confidence level, we can proceed in the following procedural steps in order to conduct this test first. Is it appropriate to the normal distribution? Yes, it is. Because both N P and Q P a greater than five. What hypotheses are we testing? We are testing whether or not the null hypothesis P equals 0.12 And the alternative hypothesis P is less than 0.12 meaning we're conducting a one tailed test. See, let's compute P hot and the test statistic P hot is simply are over N equals 0.77 The past statistic is P hat minus P over route PQ over end. As we've written in the formula right? Which reduces down to Z equals negative 1.93 Next we compute the P value. We use Z table to identify that the area under the normal curve left of the Z score is .0268. As we illustrated the graph on the right. Do we use this information to redact? H? Not? No, we do not because P is greater than alpha and we interpret this finding to suggest that we lack evidence that P is less than .12.

We have a sample with N equals 83. And of that 83 members of the sample 64 satisfy a certain condition. Want to meet our equal 64. Now we want to use the sample data to test the claim that p does not equal 2.75 for the population at a confidence level of 5% or alpha equals 0.5 Now that we've identified the confidence level, we can go to the following procedural steps to conduct this hypothesis test first. Is the normal distribution appropriate to use? Yes, it is. Because both N. P and Q. P. R greater than five. B. One of the hypotheses were testing we're testing whether or not H not P equals 10.75 or H a p does not equal 0.75 Ak We're conducting a two tailed test. Next compute P hot. And the test statistic P hot is all over. N equals 20.77 Z is given by the formula on the right, taking in as input P hat P Q n N producing down to Z equals 0.42 Next we compute the P value. We use the table to see the area outside Z equals plus and minus 0.42 as we shaded in yellow and the graph on the right. This P value corresponds to 0.6745 Next we reject H not, no, we do not. P is greater than alpha and we interpret this signing to suggest that we lack evidence that P does not equal .75.

All right. We have a sample of size and equals 196. And in that sample we see 29 objects satisfying a certain observation we want to observe that is R equals 29 out of 10 equals 196. We want to use this data to test the claim that P is greater than 092 with alpha equals 0.5 Or confidence level 5%. Now that we've identified the confidence level, we can proceed in the following procedural steps in order to conduct this hypothesis test first. Is it appropriate to use the normal distribution? Yes, it is. Because N. P and Q. P. R both greater than five secondly what hypotheses are retesting? We're testing H and R P equals 50.92 H. A. P greater than 0.92 Which means we're conducting a right tailed or one tailed test. Next compute P. A. And the test statistic he had simply are over end or 0.148 plugging that as well as P. Q and N. Into rz stat formula on the right. Give Z equals 0.271 or 2.71 next let's compute the P value based on our Z equals 2.71 We can use this table to identify the P value. The P value is simply the area under the normal curve to the right of the Z score. Since this is the right tool test from the table, we get P equals 0.34 We've illustrated this in the graph on the right next. We use this P value to reject H. Not. Yes, we do because he is the alpha. And we interpret this to mean that we have evidence that P is greater than 0.92

From 26 question a probability off A and the is equal toe for built off. A times for ability off me is in a so this is equal to probability of A, which is probability off 65 plus times for ability off flu given 65 plus, which is able to 12.5% times 14% which is equal toe at 1.75%. Could you be, um, the probability that it is 65 negative is equal culpability that is one minus programs off 65 politics, which is equal toe one minus swearing 10.5% which is equal toe eight 7.5% conflict. So therefore, will it own 65 negative and you is equal toe probability off 65 Negative. Which is which is 4.875 times probability flu, which is 4.24 which is equal toe on 21%. Or offering 21 which can see, um, question. See four questions for Coach Andy again. Um um, probability off 65 plus and flu is equal to profit off 65 plus times appropriate or flu given 65 plus, which is equal toe Oh 0.95 times work, open plan and for could soon be probability off 65 negative and blue is equal to profit off 65 negative, which is open toe or five times probability off flu given 65 negative, which is open to four. So it's if we're to a 40.12 question the again coach in a project on 65 plus and the flu is equal to, uh, open five times opening 14 which is equal to a one point or seven and for good can be for Will tee off 65 negative and through she is equal to open five times open to four. She's all 0.112


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