Assume that we want to construct a confidence interval. Do one of the following, as appropriate(a) find the critical value t_{2/2} (b) find the critical value z a/2 or () state that neither the normal distribution nor the t distribution applies . Here are summary statistics for randomly selected weights of newborn girls: n = 211 , overline x =30.7 hg; s = 6.6hg . The confidence level is 99% ?
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Assume that we want to construct a confidence interval. Do one of the following, as appropriate(a) find the critical value t_{2/2} (b) find the critical value z a/2 or () state that neither the normal distribution nor the t distribution applies . Here are summary statistics for randomly selected weights of newborn girls: n = 211 , overline x =30.7 hg; s = 6.6hg . The confidence level is 99% ?
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- = = A dietician selected a random sample of n 50 male adults and found that their average daily intake of dairy products was x 756 grams per day with a standard deviation of s 35 grams per day. Use this sample information to construct a 95% confidence interval for the mean daily intake of dairy products for men.Dr. Moore wanted to estimate the mean cholesterol level for all adult men living in Hartford. He took a sample of 25 adult men from Hartford and found that the mean cholesterol level for this sample is 186 mg/dL with a standard deviation of 12 mg/dL. Assume that the cho- lesterol levels for all adult men in Hartford are (approximately) normally distributed. Con- struct a 95% confidence interval for the population mean u.There are 200 60-year old women with glaucoma that had their blood pressures (BP) taken. The mean BP was 140 mmHg and the SD was 25mmHg. What is the 95% lower confidence interval estimate for the population's mean BP of 60-year old women with glaucoma? ( 5pts) Note: SEM = SD/ vn Do not put mmHg. Use 1 decimal places. Give the lower confidence limit value only. Answer:
- Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value ta/2. (b) find the critical value Za/2, or (c) state that neither the normal distribution nor the t distribution applies. Here are summary statistics for randomly selected weights of newborn girls: n= 163, x= 29.3 hg, s= 7.7 hg. The confidence level is 95%. @ Select the correct choice below and, if necessary, fill in the answer box to complete your choice. OA 1/2 = W S Wiew an example дв ad OB. Za/2= (Round to two decimal places as needed.) (Round to two decimal places as needed.) OC. Neither the normal distribution nor the t distribution applies. X अवा # 3 E Get more help. F3 D 80 C $ 4 9 R F % 5 V F T G ^ 6 B MacBook Air Y H 7 x U N * 30 8 J DII FO I M ( 9 K F9 O Л. : ; I { 11 [ Check answer ? command option | 1 } 1 deleThe 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 thread manufacturer wants to estimate the mean tensile strength of a particular type of thread. A random sample of 15 examples of the thread are tested. The mean tensile strength of the sample is 7.64 pounds with a sample standard deviation of 0.48 pounds. Construct a 95% confidence interval for the mean tensile strength of that type of thread. In the next question, you will enter the upper limit of the interval. For now, enter the lower limit of the interval, rounded to two decimal places.
- Do male and female servers work the same number of hours? A sample of 29 female servers worked an average of 23 hours per week, with a standard deviation of 4. A sample of 16 male servers worked an average of 24 hours per week, with a standard deviation of 2. Construct a 95% confidence interval for the mean difference in hours worked for male and female servers. You may assume that the populations are normal. Lower bound on CI = (use two decimal places) Upper bound on CI = (use two decimal places)I want to construct a confidence interval (CI) for estimating the population mean of IQ scores. I sampled 65 university students and measured their IQ. For these 100 participants, the sample mean and standard deviation are 102 and 14.8, respectively. Compute the 95% CI and interpret the CI. Show your work (e.g., R code and/or equation) if you want to get potential partial marks in case that your final answer is wrong.The mg of nicotine in menthol cigarettes is normally distributed and a sample of 25 menthol cigarettes had a standard deviation of 0.24 mg of nicotine. We will be constructing a 90% confidence interval containing the population variance. The left hand critical value is . Round to 3 decimal places. The right hand critical value is . Round to 3 decimal places. The true variance of the mg of nicotine found in menthol cigarettes is in the interval ( , ). Round your answers to 3 decimal places.
- The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. 50 newborn elephants are weighed. The sample mean is 244 pounds. a/ Determine the point estimate b/ We should use a z-test or a t-test? Explain the reason you choose z-test or a t-test. c/ Find the margin of error (error bound EBM) for a 95% confidence interval for the population mean weight of newborn elephants. (You must use the formula for the margin of error EBM to solve this question. Don’t calculate the confidence interval). Round your answer to 4 decimal places.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…calculate t value if df is 24 and 95% confidence interval