QUESTION 4 Which of the following P-values would likely lead to the rejection of the null hypothesis. 12% O 30% 95% 4%
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Q: a recent Super Bowl, a TV network predicted that 42 % of the audience would express an interest in…
A: given data, p=0.42n=100x=37p^=xn=37100=0.37we have to find out test statistica value for the…
Q: Ho: p = 0.71 H₁:p 0.71
A: The hypothesis is H0: p=0.71 H1: p≠0.71
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A: Given, p^1=0.74p^2=0.59
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A: The null and alternative hypotheses are stated below:
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Q: b. The null and alternative hypotheses would be: Ho: ? (please enter a decimal) H1: ? C (Please…
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Q: In a recent Super Bowl, a TV network predicted that 21 % of the audience would express an interest…
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Q: In a recent Super Bowl, a TV network predicted that 39 % of the audience would express an interest…
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Q: In a recent Super Bowl, a TV network predicted that 35 % of the audience would express an interest…
A: Given: Hypothesis: H0:p=0.35H0:p=0.35Ha:p>0.35Ha:p>0.35 Population, n = 123 Sample size, x =…
Q: According to the most recent General Social Survey, 70% of adults watch TV nightly. A researcher…
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A: here In a recent Super Bowl, a TV network predicted that 54 % of the audience would express an…
Q: In a recent Super Bowl, a TV network predicted that 28 % of the audience would express an int in…
A: Given data, p=0.28 p`=x/n =41/129 =0.32 Test statistic=?
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A: Given n=53 x=33
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- In a recent Super Bowl, a TV network predicted that 53 % of the audience would express an interest in seeing one of its forthcoming television shows. The network ran commercials for these shows during the Super Bowl. The day after the Super Bowl, and Advertising Group sampled 93 people who saw the commercials and found that 47 of them said they would watch one of the television shows.Suppose you are have the following null and alternative hypotheses for a test you are running:H0:p=0.53Ha:p<0.53Calculate the test statistic, rounded to 3 decimal places z=Please answer B, C, & D.The following is used for questions 25, 26, and 27. In a random sample of 200 adults, 54 say they are in favor of outlawing cigarettes. Let p be the proportion of all adults who are in favor of outlawing cigarettes. One is interested for the following hypotheses: Ho : p = 0.23 versus H. : p #0.23. 25. The standardized test statistics is about (a) -1.34. (b) -1.27 (c) 1.27 (d) 0.27 (e) 1.34 26. The P-value of this test is (a) 0.0901 (b) 0.1802 (c) 0.9099 (d) 0.8980 (e) 1.8198 27. With the significance level a of 0.01, find the rejection region.
- Question #5 and #6 only.Which of the following could be a null hypothesis for a hypothesis test? Group of answer choices p = .64 p ≠ .64 p < .64 p > .64In an experiment to study the relationship between hypertension (high blood pressure) and smoking status the following data was collected on a random sample of 180 smokers. a. Is this a Purposive or a Naturalistic study? b. State the appropriate hypotheses and test. State your conclusion. Smoking Status Light Smoker Moderate Smoker Heavy Smoker 16 Low 30 10 Blood Medium 27 20 14 Pressure High 24 18 21
- Question 3a Shoppers were asked where they do their regular grocery shopping. The table below shows the responses of the sampled shoppers. We are interested in determining if the proportions of females in the two categories are different from each other. Grocery Chain Discount Store Male 150 80 60 Female 30 a) State the null and alternative hypotheses to be tested.At a high school debate tournament, half of the teams were asked to wear suits and ties and the rest were asked to wear jeans and T-shirts. The results are given in the table below. In order to test the claim at the 0.05 level that the proportion of wins is the same for teams wearing suits as for teams wearing jeans, what would the null hypothesis be? Win Loss Suit 22 28 T-Shirt 28 22 O A. The proportions of wins is different for teams wearing suits as for teams wearing jeans. O B. The proportions of wins is the same for teams wearing suits as for teams wearing jeans. O C. The mean number of wins is the same for teams wearing suits as for teams wearing jeans. O D. The mean number of wins is different for teams wearing suits as for teams wearing jeans.According to the most recent General Social Survey, 70% of adults watch TV nightly. A researcher believes that the proportion of college students who watch TV nightly is less than this. To test this claim the research samples 400 college students and finds that 276 watch TV nightly. At the 0.05 level of significance does this data provide evidence to suggest that the proportion of college students who watch TV nightly is less than 70%? Select the correct conclusion for the hypothesis test: Question 12 options: Fail to reject the null hypothesis because the p-value is less than the level of significance. Fail to reject the null hypothesis because the p-value is greater than the level of significance. Reject the null hypothesis because the p-value is less than the level of significance. Reject the null because the p-value is greater than the level of significance.
- Males and females were asked what they would do if they received a $100 bill in the mail that was addressed to their neighbor, but had been incorrectly delivered to them. Of the 70 males sampled, 55 said yes they would return it to their neighbor. Of the 130 females sampled, 120 said yes. The overall question is: Is this sufficient evidence to say that the two true proportions are different? Choose the best null and alternative hypothesis for this test. O Null: (p1 - p2) = 0 vs Alternative (p1 - p2) > O. O Null: (p1 - p2) = 0 vs Alternative (p1 - p2) not = 0. O Null: (mean 1 - mean 2) = 0 vs Alternative (mean 1 - mean 2) not = 0. %D %3D O Null: (p1 - p2) = 0 vs Alternative (p1 - p2) < 0. %3D53% of students entering four-year colleges receive a degree within six years. Is this percent larger than for students who play intramural sports? 161 of the 283 students who played intramural sports recelved a degree within six years. What can be concluded at the level of significance of a = 0.05? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: ? ♥ Select an answer ♥ (please enter a decimal) H: ?♥ Select an answer ♥ |(Please enter a decimal) C. The test statistic|? ♥ (please show your answer to 3 decimal places.) d. The p-value = | (Please show your answer to 4 decimal places.) e. The p-value is ? ♥ a f. Based on this, we should| Select an answerv the null hypothesis. g. Thus, the final conclusion is that ... The data suggest the populaton proportion is significantly larger than 53% at a = 0.05, so there is sufficient evidence to conclude that the population proportion of students who played intramural sports who received a…spam. You've heard that at least 65% of the email you recieve through a certain email account is marked You suspect this is incorrect, so you keep track of the emails you receive on that account for the next week. In that time, you received 57 emails and 29 of them were marked as spam. State the null and alternative hypotheses of this test. Null Hypothesis (Ho): Alternative Hypothesis (HA): p? P? V To test our hypothesis, we will assume that p = Is the success-failure condition of the Central Limit Theorem satisfied? O No. Either np a. O No. The result is not significant because the p-value isSEE MORE QUESTIONS