The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives. A 95% upper confidence bound on the fraction of defective circuits is closest to

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives. A 95% upper confidence bound on the fraction of defective circuits is closest to

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Suppose we are testing brake light reaction times for a new brake light design. The current design (we will call it the old design) is known to have an average reaction time of 0.8 seconds. A random sample of 50 drivers are tested on the new design, and the 50 drivers have an average reaction time of 0.76 seconds, with a 95% confidence interval of (0.66, 0.86) seconds. Write a few sentences to explain how we should interpret this data. Explain what the null hypothesis is here, whether we rejected the null, or failed to reject, and also what the implications are.
A doctor claims that the mean number of hours of sleep that seniors in high school get per night differs from the mean number of hours of sleep college seniors get per night. To investigate, he selects a random sample of 50 high school seniors from all high schools in his county. He also selects a random sample of 50 seniors from the colleges in his county. He constructs a 95% confidence interval for the true mean difference in the number of hours of sleep for seniors in high school and seniors in college. The resulting interval is (0.57, 1.25). Based upon the interval, can the doctor conclude that mean number of hours of sleep that seniors in high school get per night differs from the mean number of hours of sleep college seniors get per night? yes because 1 is in the confidence interval yes because 0 is not in the confidence interval no because 1 is in the confidence interval no because 0 is not in the confidence interval
4) The trade magazine QSR routinely checks the drive-through service times of fast-food restaurants. A 90% confidence interval that results from examing 607 customers in Taco Bell's drive-through has a lower bound of 161.6 seconds and an upper bound of 164.7 seconds. What does this mean?
Women athletes at a certain university have a long-term graduation rate of 67%.Over the past several years, a random sample of 36 woman athletes at the school showed that 21 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from the university is now less than 67%? Use a 10% level of significance
In a certain school district, it was observed that 30% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 107 out of 295 students are only children. The school district administrators want to know if the proportion of only children in the special program is significantly different from the proportion for the school district. Test at the a = 0.01 level of significance. What is the hypothesized population proportion for this test? p = (Report answer as a decimal accurate to 2 decimal places. Do not report using the percent symbol.) Based on the statement of this problem, how many tails would this hypothesis test have? O one-tailed test O two-tailed test Choose the correct pair of hypotheses for this situation: Но:р 3 0.3 Но :р — 0.3 | Но:р — 0.3 (A) (B) (C) H:p 0.3 Но: р— 0.363| Но :р — 0.363 | Но: р— 0.363 (D) (E) (F) H:p 0.363 (В) (D) (E) (F) Using the normal approximation for the…
A university administrator wishes to know if there is a difference in average time for female and male graduates from Universiti Selangor to get a job after completing a study. Therefore, 17 female and 12 male fresh graduates are surveyed, and the results are summarized in Table 2. At the 5% level of significance, test whether the data provide sufficient evidence to conclude that the average time to get job after graduates for female is longer than male. b. s2 Female 17 3.2 34 Male 12 2.4 44 Table 2 (10 marks)
In a recent study, two treatments were compared to see which one helped symptoms of migraines. Of the 100 people who had treatment 1, 25 improved. Of the 100 people who received treatment 2, 39 improved. Is there a statistically significant difference in the proportions of patients who show improvement between treatments 1 and 2. Run the test at a 5% level of significance. 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value.
Reports show that 92% of online searches in March 2021 use Google as the search engine. A network administrator wants to determine whether the percentage of searches that use Google is different in their company. They sample 800 searches and find that 758 of them use Google. Can they conclude that the percentage of searches that use Google in this company differs from 92%? Use the a = 0.05 level of significance.
2. It has been hypothesized that treatments of plastic used in optic lenses will improve wear. Four different treatments are to be tested. To determine whether any differences in mean wear exist among treatments, 28 castings from a single formulation of the plastic were made and 7 castings were randomly assigned to each of the treatments. Wear was determined by measuring the increase in "haze" after 200 cycles of abrasion (better wear being indicated by smaller increases). The data collected are reported in the accompanying table. Treatment BcD A 9.16 11.95 11.47 11.35 13.29 15.15 9.54 8.73 12.07 14.75 11.26 10.00 11.97 14.79 13.66 9.75 13.31 15.48 11.18 11.71 12.32 13.47 15.03 12.45 11.78 13.06 14.86 12.38 83.90 98.65 87.00 76.37 11.986 14.093 12.429 10.910 Is there evidence of a difference in mean wear among the four treatments? Use a= 0.05 [HINT: Total SS = 97.0319 and SST = 36.7497] 2.1 Estimate the mean difference in “haze" increase between treatments B and D, using a 99%…
According to the National Health and Nutrition Examination Survey (NHANES) sponsored by the U.S. government, a random sample of 712 males between 20 and 29 years of age and a random sample of 1,001 males over the age of 75 were chosen and the weight of each of the males were recorded (in kg). Do the data provide evidence that the younger male population weighs more (on average) than the older male population? Use “Y” for ages 20-29 and “S” for ages 75+. It was found that x̅Y=83.4, sY=18.7, x̅S=78.5, and sS=19.0. a) Is the normality condition met for this test? (see picture) No, since the histograms show skewness and outliers . Yes, since both sample sizes are larger than 30. No, since both sample sizes, nY and nS, are less than 30. Yes, since it was stated that the two populations are normally distributed. b)What are the correct degrees of freedom for this test? Group of answer choices 999 1000 1712 1711 711
An accountant believes that the percentage of accounts that will be uncollectible increases as the ages of the accounts increase. Here the age of an unpaid account is the number of days elapsed since the invoice date. To test this theory, the accountant randomly selects independent samples of 500 accounts with ages between 31 & 60 days and 500 accounts with ages between 61 & 90 days from the accounts receivable ledger dated one year ago. When the sampled accounts are examined, it is found that 10 of the 500 accounts with ages between 31 & 60 days were eventually classified as "uncollectible", while 27 of the 500 accounts with ages between 61 & 90 days were eventually classified as "uncollectible". Let pi = proportion of accounts with ages between 31 & 60 days and p2 = proportion of accounts with ages between 61 & 90 days.
A certain type of battery supercharger was found to take a mean of 15.4 minutes to charge a standard electric vehicle (EV) battery. A firmware update for this type of supercharger was recently released, but the update has received negative reviews from users. A researcher suspects the mean time the updated superchargers take to charge a standard EV battery is greater than 15.4 minutes. To test his claim, he chooses 15 of the updated superchargers at random and measures the time it takes each one to charge a standard EV battery. He finds that the superchargers take a sample mean of 16.1 minutes to charge a standard EV battery, with a sample standard deviation of 1.8 minutes. Assume that the population of amounts of time to charge a standard EV battery using the updated superchargers is approximately normally distributed. Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.05 level of significance, to support the claim that u, the mean time…