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- In a random sample of 850 consumers. 385 reported that they were able to purchase a Playstation 5. Which of the following is a 98% confidence interval for the population proportion of consumers that were able to purchase a Playstation 5? A) (0.4131, 0.4927) B)(0.4089, 0.4969) C)(0.4248, 0.4810) D) (0.4194, 0.4864)Construct the 95% confidence interval for the difference in weight loss between the low fat and the low carbohydrate diets. Low Fat Low Carbohydrate x¯1=5.9lbs1=3.2lbn1=63 x¯1=2.6lbs1=5.9lbn1=79 Group of answer choices −0.7lb<μ<4.9lb Not enough information 0.4lb<μ<3.2lb No answer text provided. 1.8lb<μ<4.8lb None of theseAn ecologist who analyzes water samples tests the nullhypothesis that any contaminants in the water are belowdangerous concentrations. Because he uses a = 0.05, aset of samples from a small lake that produced a P-valueof 0.07 led him to conclude that the evidence did notpoint to unsafe water conditions. Which is true?A) There’s a 7% chance the lake’s water really is safe.B) There’s a 93% chance the lake’s water really is safe.C) There’s a 7% chance his sampling would have shownas much contamination as it did even if the lake’swater really is safe.D) If the lake’s water really is unsafe, there’s a 5%chance he wouldn’t notice.E) If he had taken more samples, he probably wouldhave rejected the null and concluded that the waterwas unsafe.
- In testing hypotheses, which of the following would be strong evidence against the null hypothesis?a. Obtaining data with a test statistic of small magnitude.b. Obtaining data with a small P-value.c. Obtaining data with a large P-value.d. Using a large level of significance, α.e. Using a small level of significance, αA random sample of students at a college shows that 54 of 200 students had part-time jobs. Which of the follow-ing is the correct formula for a 90% confidence interval for the proportion of all students at this college withpart-time jobs?A) 0.27 { 1.28B10.27210.732200B) 0.27 { 1.28B10.5210.52200C) 0.27 { 1.645B10.27210.732200D) 0.27 { 1.645B10.5210.52200E) 0.27 { 1.96B10.27210.732200Researchers investigating the new Johnson & Johnson COVID Vaccine selected a random sample of 200 individuals and gave them the vaccine. Of those, 52 indicated they were experiencing side effects from the vaccine. If 5,000 people took the drug, which of the following is closest to the interval estimate of the number of people who would indicate they were experiencing side effects from the vaccine at a 95 percent confidence level? a (940, 2750) b (995, 1605) c (0.199, 0.321) d (0.209, 0.311) e (1045, 1555)
- The following 3x3 contingency table contains observed frequencies for a sample of 240. Column variable row variable A B C P 20 30 20 Q 30 60 25 R 10 15 30 a-State null and alternative hypotheses totest for independence of the row and column variables. b-What are expected frequenciesfor each item? c-What is the value of test statistic? d-Using alpha= 0.05.What is the degree of freedom? What is the p-value? Using the p-value approach, what is your conclusion?Consider a sample of 48football games, where 31 of them were won by the home team. Use a 0.01 significance level to test the claim that the probability that the home team wins is greater than one-half. Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. H0: p=0.5 H1: p>0.5 B. H0: p=0.5 H1: p≠0.5 C. H0: p=0.5 H1: p<0.5 D. H0: p>0.5 H1: p=0.5 Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is. (Round to two decimal places as needed.) Identify the P-value for this hypothesis test. The P-value for this hypothesis test is. (Round to three decimal places as needed.) Identify the conclusion for this hypothesis test. A. Reject H0. There is not sufficient evidence to support the claim that the probability of the home team winning is greater than one-half. B. Reject H0. There is sufficient evidence to…For a population data set sigma = 12.7 How large a sample should be selected so that the margin of error of estimate for a 96 confidence interval for u is 3.20?
- A national opinion poll found that 44% of all American adults agree that parents should be given vouchers that are good for education at any public or private school of their choice. The result was based on a small sample. How large of a sample is required to obtain a margin of error of 0.03 in a 99% confidence interval? a. Use the national opinion poll from a previous study to find the sample size: b. Use the conservative guess of p = 0.5 to calculate the sample size:A researcher used a repeated-measures experiment to examine the effectiveness of motivational workshop on work productivity. If the data collected in a sample of n = 30 employees produced the dependent sample t-test statistic t = 2.09. Which of the following the correct decisions regarding the null hypothesis, H0 for the 2-tails tests?Which of these cells would correctly find a bootstrapped 95% confidence interval for the mean weight of a population? a. orig_sample = Table().read_table("Sample.csv") new_means = make_array() for i in np.arange(1000): new_mean = np.average(orig_sample.sample().column("Weight")) new_means = np.append(new_means, new_mean) Lower_bd = percentile(2.5, new_means)Upper_bd = percentile(97.5, new_means) print(f"The 95% confidence interval is from {Lower_bd} to {Upper_bd}") b. orig_sample = Table().read_table("Sample.csv") new_means = make_array() for i in np.arange(1000): new_mean = np.average(orig_sample.sample().select(Weight)) new_means = np.append(new_means, new_mean) Lower_bd = percentile(5, new_means)Upper_bd = percentile(95, new_means) print(f"The 95% confidence interval is from {Lower_bd} to {Upper_bd}") c. orig_sample = Table().read_table("Sample.csv") new_means = make_array() for i in np.arange(1000): new_mean = np.average(orig_sample.column(Weight))…
