places.90%????95%????96%97%????98%????99%2. For a data set obtained from a random sample, n = 81 and x = 48.25. It is known???????that ? = 4.8.What is the point estimate of ?? Round to two decimal placesMake a 95% confidence interval for ?. What is the lower limit? Round to two decimal places.Make a 95% confidence interval for ?. What is the upper limit? Round to two decimal places.What is the margin of error of estimate for part b? Round to two decimal places.3. Determine the sample size (nfor the estimate of ? for the following.E = 2.3,?? = 15.40, confidence level = 99%. Round to the nearest whole number.E = 4.1,?? = 23.45, confidence level = 95%. Round to the nearest whole number.E = 25.9,?? = 122.25, confidence level = 90%. Round to the nearest whole number.4. True or False.a.The null hypothesis is a claim about a population parameter that is assumed to be false until it is declared false.A.???TrueB.???Falseb. An alternative hypothesis is a claim about a population parameter that will be true if the null hypothesis is false.A.???TrueB.???Falsec. The critical point(s) divide(s) is some of the area under a distribution curve into rejection and nonrejection regions.A.???TrueB.???Falsed. The significance level, denoted by ?, is the probability of making a Type II error, that is, the probability of rejecting the null hypothesis when it is actually true.A.???TrueB.???Falsee. The nonrejection region is the area to the right or left of the critical point where the null hypothesis is not rejected.A.???TrueB.???False5. Fill in the blank. The level of significance?in a test of hypothesis is the probability of making a?________. It is the area under the probability distribution curve where we reject?H0.A.???Type I errorB.???Type II errorC.???Type III error6. Consider H0: ? = 45 versus H1: ? < 45. A random sample of 25 observations produced a sample mean of 41.8. Using ? = .025 and the population is known to be normally distributed with ? = 6.What is the value of z? Round to two decimal places.Would you reject the null hypothesis?Reject HoDo not reject Ho7. The following information is obtained from two independent samples selected from two normally distributed populations.n1 = 18???????????x1 = 7.82?????????1 = 2.35n2 =15???????????x2 =5.99??????????2 =3.17A. What is the point estimate of ?1 ? ?2? Round to two decimal places.B. Construct a 99% confidence interval for ?1 ? ?2. Find the margin of error for this estimate.??????Round to two decimal places.8. The following information is obtained from two independent samples selected from two???????????populations.n1 =650??????????x1 =1.05?????????1 =5.22n2 =675??????????x2 =1.54?????????2 =6.80Test at a 5% significance level if ?1 is less than ?2.a)??Identify the appropriate distribution to use.t distributionnormal distributionb)?What is the conclusion about the hypothesis?A.???Reject HoB.???Do not reject Ho9. Using data from the U.S. Census Bureau and other sources, www.nerdwallet.com estimated that considering only the households with credit card debts, the average credit card debt for U.S. house- holds was \$15,523 in 2014 and \$15,242 in 2013. Suppose that these estimates were based on random samples of 600 households with credit card debts in 2014 and 700 households with credit card debts in 2013. Suppose that the sample standard deviations for these two samples were \$3870 and \$3764, respectively. Assume that the standard deviations for the two populations are unknown but equal.a)????Let ?1 and ?2 be the average credit card debts for all such households for the years 2014 and 2013, respectively. What is the point estimate of ?1 ? ?2? Round to two decimal places. Do not include the dollar sign.b)????Construct a 98% confidence interval for ?1 ? ?2. Round to two decimal places. Do not include the dollar sign.What is the lower bound? Round to two decimal places.What is the upper bound? Round to two decimal places.c)????Using a 1% significance level, can you conclude that the average credit card debt for such households was higher in 2014 than in 2013? Use both the p-value and the critical-value approaches to make this test.A.???Reject HoB.???Do not reject Ho10. Gamma Corporation is considering the installation of governors on cars driven by its sales staff. These devices would limit the car speeds to a preset level, which is expected to improve fuel economy. The company is planning to test several cars for fuel consumption without governors for 1 week. Then governors would be installed in the same cars, and fuel consumption will be monitored for another week. Gamma Corporation wants to estimate the mean difference in fuel consumption with a margin of error of estimate of 2 mpg with a 90% confidence level. Assume that the differences in fuel consumption are normally distributed and that previous studies suggest that an estimate of?sd=3sd=3?mpg is reasonable. How many cars should be tested? (Note that the critical value of?tt?will depend on?nn, so it will be necessary to use trial and error).Solution details:STATUS Answered QUALITY Approved ANSWER RATING This question was answered on: Apr 19, 2020 PRICE: \$15 Solution~000.zip (25.37 KB) Buy this answer for only: \$15 This attachment is locked × Please Enter The Email Where You Want To Receive Solution. Get this solution for only: \$

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