in a random sample of 400 Students 332 stated that there is were nonsmokers. Based on tr Sample, Compute a 90% confidence interval for the proportion of all Students who are non smokers. Then find the lower limit and upper limit of the 90%. von fatnie interval. Lower limit= Upper limit= Carry the intermediate Computation. to at least three decimal places Bound answers to two decimal Places
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- 10q3r1 (4) Body weight Bench press 148 145 176 148 154 133 189 156 181 166 217 174 210 168 150 139 137 109 151 119 172 138 219 167 142 131 143 119 151 164 136 124 149 147 129 134 219 162 169 140 149 180 187 156 198 149 156 131 220 149 154 138 212 147 122 122 133 163 136 136Suppose a certain species bird has an average weight of x = 3.2 grams. Based on previous studies we can assume that the weights of these birds have a normal distribution with sigma = 0.34 grams. For a small group of 18 birds , find a 98% confidence interval for the average weights of these birds. Write answer in interval notation with two decimal places.In a sample of 470 adults, 306 had children. Construct a 99% confidence interval for the true population proportion of adults with children.Give your answers as decimals, to three places < p <
- Construct a confidence interval for p, - Pp2 at the given level of confidence. X1 = 384, ng = 508, x2 = 445, n2 = 573, 90% confidence The researchers are % confident the difference between the two population proportions, p, - P2, is between and (Use ascending order. Type an integer or decimal rounded to three decimal places as needed.)С. Consider another brand C of car tyres. From a random sample of 36 car tyres, it has been found that 26 of these car tyres last longer than 40,000 km. Find a 95% confidence interval corresponding to the sample proportion of brand C car tyres which last longer than 40,000 km. Give values correct to three decimal places.A psychologist is interested in the mean IQ score of a given group of children. It is known that the IQ scores of the group have a standard deviation of 15. The psychologist randomly selects 80 children from this group and finds that their mean IQ score is 105. Based on this sample, find a 99% confidence interval for the true mean IQ score for all children of this group. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: ? Upper limit:
- A psychologist is interested in the mean IQ score of a given group of children. It is known that the IQ scores of the group have a standard deviation of 14. The psychologist randomly selects 80 children from this group and finds that their mean IQ score is 111. Based on this sample, find a 99% confidence interval for the true mean IQ score for all children of this group. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: Upper limit:Forty-seven items are randomly selected from a population of 540 items. The sample mean is 36, and the sample standard deviation 9. Develop a 99% confidence interval for the population mean. (Round the final answers to 2 decimal places.) The confidence interval is between andIn a random sample of 325 students at a university, 269 stated that they were nonsmokers. Based on this sample, compute a 90% confidence interval for the proportion of all students at the university who are nonsmokers. Then complete the table below. Carry your intermediate computations to at least three decimal places What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval?
- A graduate Statistic Student wants to estimate the true proportion of county college ("c.c.") students in here state who go on to receive an Associate Degree (AD ), She is able to take a random sample of n = 100 of those c.c. stuents, and 64 of those students received an AD. Find her 90% confidence interval for the true proportion of c.c. students in her state that receive an AD. Be sure to either i)Use full calculator syntax including proper label on your output or i)write down the formula, plug in the appropriate nun bers, and then give the calculated interval including a proper label. 1-prop-Z Int x: 64,n: 100, Clevel.90, interval (.56105,.71895 b. 1-prop-Z Int x: 64,n: 100, Clevel .90, interval (.66105,.71896 1-prop-Z Int x: 64,n: 100, Clevel .90, interval (.51105,.70950 1-prop-Z Int x: 64,n: 100, Clevel .90, interval (.61105,.81105 by C.In a sample of 300 adults, 198 had children. Construct a 90% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places RectanIn a sample of 360 adults, 266 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three placesRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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