If a treatment is expected to increase scores among a population with u= 20, then the alternative hypothesis is µwith treatment > 20.
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- A random sample of 60 trials resulted in 18 successes. List the null and alternate hypothesis if you were to test the claim that the population proportion exceeds 20% at an ?=0.01.10% of all Americans suffer from sleep apnea. A researcher suspects that a lower percentage of those who live in the inner city have sleep apnea. Of the 369 people from the inner city surveyed, 22 of them suffered from sleep apnea. What can be concluded at the level of significance of αα = 0.01? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: Ho: ? μ p Select an answer = < ≠ > (please enter a decimal) H1: ? μ p Select an answer > ≠ < = (Please enter a decimal) The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject accept fail to reject the null hypothesis. Thus, the final conclusion is that ... The data suggest the population proportion is not significantly smaller than 10% at αα = 0.01, so…20% of all college students volunteer their time. Is the percentage of college students who are volunteers smaller for students receiving financial aid? Of the 356 randomly selected students who receive financial aid, 61 of them volunteered their time. What can be concluded at the αα = 0.10 level of significance? For this study, we should use? Select an answer z-test for a population proportion? or t-test for a population mean? The null and alternative hypotheses would be: H0:H0:? μ or p Select an answer ≥ = < > ≤ ≠ _______(please enter a decimal) H1:H1:? μ or p Select an answer ≥ ≠ = < > ≤ _______ (Please enter a decimal) 2. The test statistic ? t or z = ______ (please show your answer to 3 decimal places.) The p-value = ______ (Please show your answer to 4 decimal places.) The p-value is ? > or ≤ α 3. Based on this, we should Select an answer fail to reject? reject ? accept ? the null…
- A random sample of size 24 and mean 2.547 was selected from normal distribution if we want to test that the population mean is less than 5.547, knowing that the sample variance 25, the critical value for alpha =0.01 is? a. 2.807 b. 2.500 c. 2.547 d. 2.33Is there a difference between men and women when it comes to seeking preventative healthcare? At the a = 0.01 level of significance, test the claim that the proportion of those who have had a wellness visit with their physician in the past year is the same for men and women. Let pp represent the proportion of females who have had a wellness visit with their physician in the past year and pM represent the proportion of males who have done the same. Which would be correct hypotheses for this test? O Ho:PF = PM, H1:PF > PM O Ho:PF + PM, H1:PF > PM O Ho:PF = pM, H1:pF < pM O Ho:PF = PM, H1:PF # PM In a random sample of 331 females, 306 have had a wellness visit with their physician in the past year. In a random sample of 272 males, 237 have had a wellness visit with their physician in the past year. Find the test statistic (2 decimal places): Give the P-value (4 decimal places - if less than 0.001 answer 0): Which is the correct result: O Reject the Null Hypothesis O Do not Reject the Null…14% of all Americans suffer from sleep apnea. A researcher suspects that a lower percentage of those who live in the inner city have sleep apnea. Of the 359 people from the inner city surveyed, 43 of them suffered from sleep apnea. What can be concluded at the level of significance of αα = 0.05? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: Ho: ? μ p Select an answer = ≠ > < (please enter a decimal) H1: ? p μ Select an answer < > = ≠ (Please enter a decimal) The test statistic ? t z = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject fail to reject accept the null hypothesis. Thus, the final conclusion is that ... The data suggest the populaton proportion is significantly smaller than 14% at αα = 0.05, so…
- A pharmaceutical company is testing a new drug for reducing cholesterol levels. to approve the drug for the next round of testing, they need to show that this drug reduces mean total cholesterol level by at least 50 mg/dL. They initially plan a study that involves 50 subjects and a significance level of 0.05, but they discover that the power of the test against this effect size is only 0.24. what are two ways they can increase teh power of the test without changing effect size?Average IQ scores for young school children are known to be 100 (SD = 15). However, the literature indicates that children’s intelligence may be decreased if their mothers have German measles during pregnancy. Using hospital records, a researcher obtained a sample of n = 25 school children whose mothers all had German measles during their pregnancies. Do the data indicate that children born to mothers who had German measles during the pregnancy have significantly lower IQ scores than typical young school children? DATA (Fictional) child IQ scr 1 91 2 111 3 97 4 95 5 96 6 95 7 95 8 96 9 94 10 93 11 95 12 92 13 96 14 91 15 97 16 93 17 92 18 99 19 93 20 97 21 85 22 94 23 96 24 95 25 96 Looking at the research scenario for this lab, you will see that it says, “ literature indicates that children’s intelligence may…The standard error of most sample estimators approaches zero as the sample size increases, so almost any difference between the sample statistic and the hypothesized parameter value, no matter how tiny, will be significant if the sample size is large enough. Researchers who deal with large samples must expect “significant” effects, even when an effect is too slight to have any practical importance. Is an improvement of 0.2 mpg in fuel economy important to Toyota buyers? Is a 0.5 percent loss of market share important to Hertz? Is a laptop battery life increase of 15 minutes important to Dell customers? Such questions depend not so much on statistics as on a cost/benefit calculation. Since resources are always scarce, a dollar spent on a quality improvement always has an opportunity cost (the foregone alternative). If we spend money to make a certain product improvement, then some other project may have to be shelved. Since we can’t do everything, we must ask whether a proposed product…
- The effectiveness of surgery for weight loss reported here found that "The surgery was associated with significantly greater weight loss [than the control group who dieted] through 2 years (61.3 versus 11.2 pounds, P<0.001)." What test could have been used and how would it have been computed?A team of engineers at the research center of a car manufacturer performs crash tests to determine the proportion of times the cars' airbags fail to operate in a crash. With the airbag system's new modified design, the team expected to reduce the failed proportion to be below last year's proportion of 0.08. They decided to test HO: p = 0.08 versus Ha: p < 0.08, where p = the proportion of failed airbags during crash tests. If 300 crashes performed in the lab resulted in 18 failures, which of the following is the test statistic for this test?Suppose the national average dollar amount for an automobile insurance claim is $777.86. You work for an agency in Michigan and you are interested in whether or not the state average is less than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≥ 777.86, Alternative Hypothesis: μ < 777.86. You take a random sample of claims and calculate a p-value of 0.0232 based on the data, what is the appropriate conclusion? Conclude at the 5% level of significance. Question 9 options: 1) The true average claim amount is significantly different from $777.86. 2) The true average claim amount is significantly less than $777.86. 3) The true average claim amount is significantly higher than $777.86. 4) We did not find enough evidence to say the true average claim amount is less than $777.86. 5) The true average claim…
