1. A statistical hypothesis is a claim or assertion about a parameter or some parameters of one or more populations. А. False В. True
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- Daily anxiety was measured on a scale from 1 (not at all anxious) to 5 (very anxious) in a random sample of 2000 city dwellers from across the U.S. They found that M = 4.13, 95% CIs [4.06, 4.20].How would you interpret these results? What conclusions would you draw about the precision of the point estimate? What statistical decision would have been made in this scenario if the researchers employed Null Hypothesis Significance Testing instead of the New Stats?1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C. H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2. At 5% level, what are the critical values for testing equality of mean weights in problem 1? A. 2.18 B. -2.18 and 2.18 C. -1.78 D.-1.78 and 1.78 3.What is the best decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical…1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C. H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2.What would be the degree of freedom for the test statistic in problem 1? A. 6 B. 9 C. 12.7 D. 14 3. What would be the computed test statistic in problem 1? A. 2.93 B. 3.57 C. 8.44 D. 11.48
- 1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C. H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2. What is the best decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical region and we do not reject the null hypothesis. C. The computed test statistic falls in the critical region and we reject the null hypothesis. D.The computed…1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C.H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2. What is the best decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical region and we do not reject the null hypothesis. C. The computed test statistic falls in the critical region and we reject the null hypothesis. D. The computed test statistic does not fall…Suppose the average blood sugar level in 35- to 44-year-olds is 4.86 (mmol/L). Do sedentary people have a different blood sugar level than that of the general population? To answer that question, a hypothesis test is planned to collect data from a group of 200 sedentary people in this age group. State the null and alternative hypothesis for this study. A. H0: u (mean) ≠ 4.86 mmol/L vs Ha: u (mean) = 4.86 mmol/L B. H0: u (mean) = 4.86 mmol/L vs Ha: xbar ≠ 4.86 mmol/L C. H0: u (mean) = 4.86 mmol/L vs Ha: u (mean) > 4.86 mmol/L D. H0: u(mean) = 4.86 mmol/L vs Ha: u (mean) ≠ 4.86 mmol/L
- Suppose we take a sample of 2,500 blood donors from a population for which 50% (0.5) have type O+ blood. (a) Into what range of possible values should the sample proportion fall 95% of the time, according to the Empirical Rule? to (b) If the sample included only 625 donors instead of 2,500, would the range of possible sample proportions be wider, more narrow, or the same as with a sample of 2,500 donors? Explain your answer, and explain why it makes intuitive sense. The range would be with 625 donors compared to a sample of 2,500 donors since the standard deviation of the sampling distribution would be . This makes intuitive sense because if fewer donors are included in the sample, the proportion will be reliable as an estimate of the proportion.3) A firm in Lebanon has developed a chemical solution that can be added to car gasolinewhich they believe will increase the miles per gallon that cars will get. The owners areinterested in estimating the difference between mean mpg for cars using the chemicalsolution versus those that are not using the solution. The following data represent the mpgfor independent random samples of cars from each population.with Solution without Solution______________________________n1 = 36 n2 = 42 x1 = 25.45 x2 = 24.1 _______________________________Assume that the populations are normally distributed and the population standarddeviations are known to be σ1 = 3.95 (with solution) and σ2 = 3.09 (without solution).Given this data, can the owners believe that there is a difference between mean mpg forcars using the chemical solution versus those that are not using the solution? Test using analpha level equal to 0.05.4) Given the following null and alternative hypothesis:H0: σ 2 ≤ 52HA : σ 2 > 52and the…Suppose θb is an unbiased point estimator for a parameter θ. We obtain 10,000 different random samples and we compute the value of θb every time. Can we say that θb will underestimate θ a total of 5000 times and it will overestimate θ another 5000 times? What do you expect to see in this experiment?
- From a study from earlier this year, it is believed that approximately 7% of U.S. adults do not use the internet. Suppose researchers are doing a study to see if the proportion of Washington state adults that do not use the internet is lower than the reported U.S. adult proportion. The researchers decide to survey 200 randomly selected adults throughout the state of Washington. a. Define a parameter for hypothesis testing and write the null hypothesis and alternative hypothesis for this studyThe worker on examining his 20 rainfall sample points concluded that it is more than likely to rain on this May 1st. Describe two or three statistical steps he took to arrive at this conclusion.A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. A recent sample of 1,000 adults showed 410 indicating that their financial security was more than fair. Suppose that just a year before, a sample of 1,200 adults showed 420 indicating that their financial security was more than fair. (a) State the hypotheses that can be used to test for a significant difference between the population proportions for the two years. (Let p1 = population proportion most recently saying financial security more than fair and p2 = population proportion from the year before saying financial security more than fair. Enter != for ≠ as needed.) H0: p1−p2=0 Ha: p1−p2!=0 (b) Conduct the hypothesis test and compute the p-value. At a 0.05 level of significance, what is your conclusion? Find the value of the test statistic. (Use p1 − p2. Round your answer to two decimal places.) Find the p-value.…