a.)The null hypothesis is: . b.)The alternative hypothesis is:  c.)The test statistic is:

The manufacturer of a particular brand of tires claims they average at least 50,000 miles before needing to be replaced. From past studies of this tire, it is known that the population standard deviation is 8,000 miles. A survey of tire owners was conducted. From the 28 tires surveyed, the mean lifespan was 46500 miles. Using alpha = 0.05, can we prove that the data in inconsistent with the manufacturers claim?   We should use a     test. What are the correct hypotheses? H0:                   Ha:                    Based on the hypotheses, find the following:Test Statistic=p-value= The correct decision is to       . The correct conclusion would be:

The mean age when smokers first start is 13 years old with a population standard deviation of 1.9 years. A researcher thinks that smoking age has significantly changed since the invention of ENDS- electronic nicotine delivery systems. A survey of smokers of this generation was done to see if the mean age has changed. The sample of 30 smokers found that their mean starting age was 12.2 years old. Do the data support the claim at the 1% significance level? What are the correct hypotheses? Ho: Select an answer v ? v years H3: Select an answer v years The test is: O One-tailed O Two-tailed

A tire manufacturer claims that the mean life of its tire is 75,000 miles. The mean of a sample of 50 tires is 74,200 miles with a standard deviation of 2300 miles. Using an alpha of 5%, does the manufacturer’s claim appear to hold ture? Explain.

One way to check on how representative a survey is of the population from which it was drawn is to compare various characteristics of the sample with the population characteristics. A typical variable used for this purpose is age. The GSS 2018 found a mean age of 48.69 and a standard deviation of 17.99 for its sample of 1,495 American adults. Assume that we know from census data that the mean age of all American adults is 37.80. What is the research and the null hypotheses for a two-tailed test of means, calculate the t statistic and test the null hypothesis at the .001 significance level, and what is the decision about the null hypothesis?

Individual scores for practices towards COFLU-20 are found to be normally distributed with a mean of 3 and a standard deviation of 1.5 over the entire population. The Ministry of Health wishes to use this information to identify individuals whose score on practices towards COFLU20 is in the lower 75%, in order to give them relevant information on the best practices towards COFLU-20. You are asked to assist in calculating the score that the Ministry of Health should use to decide whether an individual should be given the relevant information?

You want to assess the differences in frequency counts for four different groups. You know there are 150 people total. Without any additional information about how the data is distributed, what would you use for the "expected frequency" in each group?     a.1/4 b.37.5 c.25 d..375

The BDHS reported that the mean cholesterol level in 2020 for all adults was 203 in a certain area. Around a sample of 3,310 participants were questioned about the cholesterol levels who attended the last month survey and it is found that mean cholesterol level is 200.3 with a standard deviation of 36.8. Is there any statistical evidence of a difference in mean cholesterol levels in that area? Conduct the test at 1% level of significance. a. Which of the following statistical tests is most appropriate in this scenario? Two-Tailed Z-Test One-Tailed Z-Test Two-Tailed T-Test One-Tailed T-Test b. What are the appropriate hypotheses for the test? Но : д — 203 Н : д + 203 Но : д + 203 Н] :д — 203 Но : и — 200.3 Ні : д > 200.3 Ho : µ = 200.3 H1 : µ < 200.3 C. After conducting the appropriate significance test, the test statistic was found to be -4.22. Therefore, the conclusion for the test is: Reject Ho, because p-value is less than 0.01 Fail to reject Ho, because -4.22 falls outside the…

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.40 ounce. You randomly select 45 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be reset? Explain your reasoning. it is that you would have randomly sampled 45 cans with a mean equal to 127.9 ounces, because it within the range of a usual event, namely within of the mean of the sample means.