1. The following data, recorded in days, represents the recovery time „for patients who are randomly treated with one of two medications to cure severe bladder infections: Medication 1 Medication 2 n1 =13 n2 =16 =20 72 =15 s =2.0 s =1.8 Find the 99% confidence interval for u1- u2, the difference in mean drug recovery times, and INTERPRET it to get a helpful conclusion about the drugs. Assume normal populations, with equal variances. As an answer to this question, enter the lower limit of the confidence interval found, rounded to two digits. For example, if the interval is [2.3923, 5.0579], enter 2.39.
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- The number of cell phones per 100 residents in countries in Europe is given in table #9.3.9 for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in table #9.3.10 also for the year 2010 ("Population reference bureau," 2013). Find the 98% confidence interval for the different in mean number of cell phones per 100 residents in Europe and the Americas. (Show all work) Table #9.3.9: Number of Cell Phones per 100 Residents in Europe 100 100 76 100 130 75 84 112 112 84 138 133 118 134 126 126 188 129 93 64 128 124 124 122 109 121 127 152 96 96 63 99 95 151 147 123 123 95 67 67 118 125 110 110 115 140 115 141 77 98 98 102 102 112 118 118 54 54 23 121 126 47 Table #9.3.10: Number of Cell Phones per 100 Residents in the Americas 158 117 106 159 53 50 78 66 88 92 42 3…In a study of morphological variation in natural populations of fruit fly, it was reported that the mean wing length of 16 females, collected in a certain area, was 4.653 mm with s = 0.012 mm; and the mean wing length of 11 females, collected in a second area, was 4.274 mm with s = 0.02. Let the distribution of the wing length be normal, find a 98% confidence interval on the ratio of the two populations’ standard deviations.Sulfur dioxide (SO2) pollution is a known cause for corrosion of exposed steel. To document this effect in actual conditions, steel weight loss (gm/m3/yr) was measured at 36 locations in a large city, along with the rate at which SO2 was deposited (mg/m2/day). Obtain an estimate of the correlation between steel loss and sulfur dioxide deposition rate and interpret it. Give a 95% confidence interval for its value. SO2 Loss 22.6 580 26.1 563 19.1 339 17.2 346 25 623 16.8 420 42.7 951 30.4 794 17.9 461 30.4 757 27.8 573 27.8 656 33 990 14.2 551 37.4 856 16.2 443 27.7 802 22.7 657 21 495 15.8 496 13.7 353 24.7 724 23.7 602 30.7 639 33.5 813 30.2 770 20.1 556 22.8 557 39.9 1001 15.5 432 40.4 940 27.9 687 39.4 919 25.1 590 28.6 640 29.7 702
- A biology student measured the ear lengths of an SRS of 10 Mountain cottontail rabbits, and an SRS of 10 Holland lop rabbits. The ear lengths for the two samples are listed in the two tables attached (see attached image). (a) Calculate a 95% confidence interval for the difference in mean ear lengths between Mountain cottontail rabbits and Holland lop rabbits. Make sure to define which is “µ1” and which is “µ2.” (You can complete the calculations either by hand or using R, but remember to show all your work.) (b) Do Holland lop rabbits have longer ears on average than Mountain cottontail rabbits? Carry out a test of significance to answer this question. Show your work at each step. Don’t forget to state the hypotheses at the start (making sure to define all parameters), and to include a conclusion in terms of the original problem. (You can complete the calculations either by hand or using R, but remember to show all your work.)In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured befor and after the treatment. The changes (before - confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness a garlic in reducing LDL cholesterol? after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.2 and a standard deviation of 19.8. Construct a 95° Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? mg/dL <µIn a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.2 and a standard deviation of 17,1. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table, Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? O mg/dLin clinical trials of allergy medicine clarinex 5 mg it was reported that 50 out of 1655 individuals in clarinex group and 31 out of 1652 individuals in placebo group experienced dry mouth as a possible side effect of their respectivetreatments. the results of a 95% confidence interval for the proption of clarinex users who ecperince dry mouth munus the proportion of placebo takers who ecpericened dry mouth are ( 0.0009,0.0220) interpret the interval in contextThe mean change in muscle thickness observed after a 4-week training regimen was 4 mm, and the 95% confidence interval (CI) for mean change was (-0.5, 8.5). Which is the best scientific conclusion for the effectiveness of this regimen ?A new cream that advertises that it can reduce wrinkles and improve skin was subject to a recent study. A sample of 68 women over the age of 50 used the new cream for 6 months. Of those 68 women, 39 of them reported skin improvement(as judged by a dermatologist). Is this evidence that the cream will improve the skin of more than 40% of women over the age of 50? Test using a= 0.05 (a) Test statistic: z= (b) The final conclusion is...A. There is not sufficient evidence to reject the null hypothesis that p=0.4 . That is, there is not sufficient evidence to reject that the cream can improve the skin of more than 40% of women over 50.B. We can reject the null hypothesis that p=0.4 and accept that p>0.4. That is, the cream can improve the skin of more than 40% of women over 50.Arsenic is a compound that occurs naturally in very low concentrations. Arsenic blood concentrations in healthy individuals are Normally distributed with a mean of 3.2 mg/dl. Some areas of the United States have naturally elevated arsenic levels in the ground and water supplies. Researchers took a sample 20 water samples from a particular area and measured the arsenic levels in each of the 20 water samples. They want to determine if there is sufficient evidence to conclude that the arsenic level in that area is higher than 3.2 mg/dl. hypothesis test for one population mean (unknown population standard deviation) confidence interval estimate for one population mean (unknown population standard deviation) hypothesis test for population mean from paired differences confidence interval estimate for population mean from paired differences hypothesis test for difference in population means from two independent samples confidence interval estimate for difference in population means from two…An agricultural researcher plants 25 plots with a new variety of yellow corn. Assume that the yield per acre for the new variety of yellow corn follows a normal distribution with unknown mean u and standard deviation o = 10 bushels per acre. If the average yield for these 25 plots is x = 150 bushels per acre. What is a 90% confidence interval for u?Research indicates that there may be a relationship between caffeinated coffee consumption and risk of depression in women. Provided is data on the amount of caffeinated coffee consumed and whether the woman was diagnosed with clinical depression. Of interest is to see if there is evidence of an association between caffeinated coffee intake and clinical depression in women. Using the provided output, what is the pvalue and what are the results of the test? Caffeinated coffee Depression <1 cup/wk >=1 cup/day 2-3 cups/day 4+ cups/day Total Yes 670 1278 564 95 2607 No 11545 22573 11726 2288 48132 Total 12215 23851 12290 2383 50739 Pearson's Chi-squared testdata: coffeeX-squared = 19.472, df = 3, p-value = 0.0002184 A. pvalue=3 <= alpha(0.05), H0 is rejected. There is an association between depression and caffeine intake in women B. pvalue=19.472 <= alpha(0.05), H0 is rejected. There is an association between depression and caffeine intake in…SEE MORE QUESTIONSRecommended textbooks for youCalculus For The Life SciencesCalculusISBN:9780321964038Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.Publisher:Pearson Addison Wesley,Calculus For The Life SciencesCalculusISBN:9780321964038Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.Publisher:Pearson Addison Wesley,