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- An article in Engineering Horizons (Spring 1990, p. 26) reported that 117 of 484 new engineering graduates were c08.indd 302 9/24/2013 6:57:11 PMSection 8-7/Tolerance and Prediction Intervals 303planning to continue studying for an advanced degree. Consider this as a random sample of the 1990 graduating class.(a) Find a 90% confidence interval on the proportion of such graduates planning to continue their education.(b) Find a 95% confidence interval on the proportion of such graduates planning to continue their education.(c) Compare your answers to parts (a) and (b) and explain why they are the same or different.(d) Could you use either of these confidence intervals to determine whether the proportion is actually 0.25? Explain your answer. Hint: Use the normal approximation to the binomial.According to a recent report, 468% of college student internships are unpaid. A recent survey of 80 college interns at a local university found that 53 had unpaid internships. a. Use the five-step p-value approach to hypothesis testing and a 0.01 level of significance to determine whether the proportion of college interns that had unpaid internships is different from 0.48. b. Assume that the study found that 46 of the 80 college interns had unpaid internships and repeat (a). Are the conclusions the same? ZStat=___ p-value=____ What is the final conclusion? The result is ____ part (a). ____ the null hypothesis. There ___ sufficient evidence that the proportion of college interns that had unpaid internships is ____ because the p-value ____ the level of significance.10.17 Refer to Exercise 10.16. Assuming that equal sample sizes will be taken from the two populations, how large a sample should be taken from each of the populations to obtain a 99% confidence interval for pA - pB with a width of at most .02? (Hint: Use p^ A = 0.3 and p^ B = .15 from Exercise 10.16.
- 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)1. A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480were delivered on time. a) Construct a 95% confidence interval for the proportion of all packages that are deliveredon time by the Postal Service.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:
- Conduct the appropriate hypothesis test and compute the p-value. Sunshine Air has determined that their no-show rate for passengers booked on a flight is 6%. The Airline has recently increased the cost of its travel insurance and and suspects the no-show rate for passengers will decrease. A random sample of 380 reservations resulted in 18 no-shows. At the 0.10 significance level, is there strong enough sample evidence to suggest that the no-show rate in less than 6%?According to a recent report, 46% of college student internships are unpaid. A recent survey of 80 college interns at a local university found that 51 had unpaid internships. a. Use the five-step p-value approach to hypothesis testing and a 0.01 level of significance to determine whether the proportion of college interns that had unpaid internships is different from 0.46. b. Assume that the study found that 42 of the 80 college interns had unpaid internships and repeat (a). Are the conclusions the same? a. Let π be the population proportion. Determine the null hypothesis, H0, and the alternative hypothesis, H1. H0: π less than< greater than> less than or equals≤ equals= greater than or equals≥ not equals≠ nothing H1: π greater than> less than or equals≤ equals= not equals≠ less than< greater than or equals≥ nothing (Type integers or decimals. Do not round.) What is the test statistic? ZSTAT= (Round to two decimal places as…According to a recent report, 46% of college student internships are unpaid. A recent survey of 80 college interns at a local university found that 51 had unpaid internships. a. Use the five-step p-value approach to hypothesis testing and a 0.01 level of significance to determine whether the proportion of college interns that had unpaid internships is different from 0.46. b. Assume that the study found that 42 of the 80 college interns had unpaid internships and repeat (a). Are the conclusions the same? b. Assume that the study found that 42 of the 80 college interns had unpaid internships and repeat (a). What is the test statistic? ZSTAT= (Round to two decimal places as needed.) What is the p-value? The p-value is (Round to three decimal places as needed.)
- 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?The Congressional Budget Office reports that 36% of federal civilian employees have a bachelor's degree or higher (The Wall Street Journal). A random sample of 115 employees in the private sector showed that 32 have a bachelor's degree or higher. Does this indicate that the percentage of employees holding bachelor's degrees or higher in the private sector is less than in the federal civilian sector? Use α = 0.05. State the null and alternate hypotheses. H0: μ = 0.36; H1: μ < 0.36H0: p = 0.36; H1: p > 0.36 H0: p = 0.36; H1: p < 0.36H0: p = 0.36; H1: p ≠ 0.36H0: μ = 0.36; H1: μ > 0.36H0: μ = 0.36; H1: μ ≠ 0.36 (b) What sampling distribution will you use? What assumptions are you making? The standard normal, since np < 5 and nq < 5.The Student's t, since np > 5 and nq > 5. The Student's t, since np < 5 and nq < 5.The standard normal, since np > 5 and nq > 5. What is the value of the sample test statistic? (Round your answer to two decimal…A random sample of 24 army recruits has a mean height of 68 inches with a standard deviation of 2.3 inches. If a 98% confidence interval is constructed, how many degrees of freedom would be allowed to find the t-critical value? a) 67 b) 97 c) 23