Question 1[2+1+3+2=8] Two brands of refrigerators, denoted A and B, are each guaranteed for 1 year. In a random sample of 100 refrigerators of brand A, 10 were observed to fail before the guarantee period ended. An independent random sample of 150 brand B refrigerators also revealed 15 failures during the guarantee period. Test the claim that 5% more of Brand A than Brand B fail before he guarantee expires. Use a significance level of 1%. (Round off to three decimals)
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- A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 76%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 76% of married couples. In a random sample of 205 married couples who completed her program, 171 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.05 level of significance?Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (For z test statistics) every expert has gotten the z test statistic wrong thus far. I have included pictures of a sample problem and formula. A. Find the value of the test statistic. (Round to three or more decimal places.) B. Find the critical value. (Round to three or more decimal places.) C. Is there enough…A student wants to see if the number of times a book has been checked out of the library in the past year and the number of pages in the book are related. She guesses that the number of times a book has been checked out of the library in the past year will be a predictor of the number of pages it has. She randomly sampled 15 books from the library, test her claim at a 0.05 level of significance. number of times checked out number of pages 13 223 45 350 49 377 30 371 3 343 7 280 17 218 7 265 20 382 23 287 1 313 30 388 20 358 2 311 40 421 The correlation coefficient: r= (round to 3 decimal places) The equation y=a+bx is: (round to 3 decimal places) y=+ x The hypotheses are: H0:ρ=0H0:ρ=0 (no linear relationship)HA:ρ≠0HA:ρ≠0 (linear relationship) (claim) Since αα is 0.05 the critical value is -2.16 and 2.16 The test value is: (round to 3 decimal places) The p-value is: (round to 3 decimal places) The decision is to reject H0H0 do not reject…Ten engineering schools in a country were surveyed. The sample contained 225electrical engineers, 80 being women; 175 chemical engineers, 40 being women. Compute a 90% confidence interval for the difference between the proportions of women in these two fields of engineering. Is there a significant difference between the two proportions? Let p1 be the population proportion of electrical engineers that are women in the schools that were surveyed and let p2 be the population proportion of chemical engineers that are women in the schools that were surveyed. The 90% confidence interval is nothing<p1−p2<nothing. (Round to three decimal places as needed.)
- Suppose a 95% confidence interval for the difference in test scores between Class 1 and Class 2 (in that order) is the following: 9 +/− 2. These results were based on independent samples of size 100 from each class. Now suppose you switch the order of Class 1 and Class 2 in your analysis but keep the data labeled correctly in terms of which class they came from. Which of the following statements is false? A) You can't do it this way. You'll get negative numbers for the difference in the means and/or negative numbers for the standard error. B) You are confident that the average for Class 2 is 7 to 11 points lower than for Class 1. C) You are still confident that the classes have significantly different mean scores. D) Your 95% confidence interval will now be entirely negative: from −7 to −11.A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 79%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 79% of married couples. In a random sample of 240 married couples who completed her program, 194 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor's claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis Ho and the alternative hypothesis H. H. :0 = 0.79 H, : o > 0.79 (b) Determine the type of test statistic to use. OSO (c) Find the value of the test statistic. (Round to three or more decimal places.) OA marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 80%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 80% of married couples. In a random sample of 215 married couples who completed her program, 180 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. A. State the null hypothesis Hoand the alternative hypothesis H1. Ho: H1: B. Find the value of the test statistic. (Round to three or more decimal places.) C. Find the critical value. (Round to three or more decimal places.) D. Is there enough evidence to support the marriage counselor's claim that the proportion of married couples for whom her program…A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 79%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 79% of married couples. In a random sample of 250 married couples who completed her program, 205 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor's claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H₁. H₂ : D H₁ : 0 (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) 0 (d) Find the p-value. (Round to three or more decimal places.) 0 (e) Is there enough…A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 77%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 77% of married couples. In a random sample of 215 married couples who completed her program, 167 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0 and the alternative hypothesis H1. H0: H1: (b) Determine the type of test statistic to use. ▼(Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the p-value. (Round to three…A statistics professor at a major university asked her students whether or not they wereregistered to vote. In a sample of 50 of the randomly sampled statistics students (from over1000 students), 35 said they were registered to vote. Find a 98% confidence interval and makea conclusion based upon those results.The Substance Abuse and Mental Health Services Administration (SAMHSA, 2017) estimates that 10.9% of the population of the United States age 18–24 had an episode of depression in the previous 12 months. Suppose a researcher takes a random sample of 225 United States adults aged 18–24. Determine the value of the sample proportion, p, such that 5% of samples of 225 United States adults age 18–24 are greater than p. You may find software or a z-table useful. Give your answer precise to three decimal places. p = IIXu and Garcia (2008)conducted a research study demonstrating that 8-month-old infants appear to recognize which samples are likely to be obtained from a population and which are not. In the study, the infants watched as a sample of n = 5 ping-pong balls was selected from a large box. In one condition, the sample consisted of 1 red ball and 4 white balls. After the sample was selected, the front panel of the box was removed to reveal the contents. In the expected condition, the box contained primarily white balls like the sample, and the infants looked at it for an average of M = 7.5 seconds. In the unexpected condition, the box had primarily red balls, unlike the sample, and the infants looked at it for M = 9.9. The researchers interpreted the results as demonstrating that the infants found the unexpected result surprising and, therefore, more interesting than the expected result. Assuming that the standard error for both means is σM = 1 second, draw a bar graph showing the two sample…A random sample of n - 236 people who live in a city were selected and 81 identified as a "dog person." A random sample of n2 - 109 people who live in a rural area were selected and 60 identified as a "dog person." Find the 90% confidence interval for the difference in the proportion of people that live in a city who identify as a "dog person" and the proportion of people that live in a rural area who identify as a "dog person." Round answers to to 4 decimal places. P1 - P2 Question Help: Message instructor Submit Question Jump to AnswerSEE MORE QUESTIONS