ASSIGNMENT Chapter 7Statistical MethodsQuestionsM & S 313-315(LARGE SAMPLES)7.10, 7.12, 7.14, 7.20, 7.22M & S 323-325(SMALL SAMPLES)7.32, 7.35, 7.40M & S 332-333(PROPORTIONS)7.50, 7.51, 7.53M & S 338-339(SAMPLE SIZE)7.66, 7.71, 7.787.10 (3 points) A random sample of 90 observations produced a mean € x = 25.9 and a standard deviation of s= 2.7.a. Find a 95% confidence interval for mb. Find a 90% confidence interval for m.c. Find a 99% confidence interval for m.Answers:7.12 (3 points) The mean and standard deviation of a random sample of n measurements are equal to 33.9 and 3.3, respectively.a. Find a 95% confidence interval for m if n= 100.b. Find a 95% confidence interval for m if n= 400.c. Find the widths of the confidence intervals you calculated in parts a and b. Whatis the effect on the width of a confidence interval of quadrupling the sample sizewhile holding the confidence coefficient fixed?7.14 (4 points) Personal networks of older adults. In sociology, a personal network is defined as the people with whom you make frequent contact. The Living Arrangements and Social Networks of Older Adults (LSN) research program used a stratified random sample of men and women born between 1908 and 1937 to gauge the size of the personal network of older adults. Each adult in the sample was asked to “please name the people (e.g., in your neighborhood) you have frequent contact with and who are also important to you.” Based on the number of people named, thepersonal network size for each adult was determined. The responses of 2,819 adults in the LSN sample yielded the following statistics on network size: € x = 14.6; s=98Sociological Methods and Research, August 2001.)a. Give a point estimate for the mean personal network size of all older adults.b. Form a 95% confidence interval for the mean personal network size of all older adults.c. Give a practical interpretation of the interval you found in part b.d. Give the conditions required for the interval in part b to be valid.Answers for 7.12 and 7.14:7.20 (2 points) Velocity of light from galaxies. Refer toe the Astronomical Journal (July1995) study of the velocity of light emitted from a galaxy in the universe, presented in Exercise 2.100(p.74). A sample of 103 galaxies located in the galaxy cluster A2142 had a mean velocity of € x = 27,117 kilometers per second (km/s) and a standard deviation of s = 1,280 km/s. Suppose your goal is to make an inference about the population mean light velocity of galaxies in cluster A2142.a. In part b of Exercise 2.100, you constructed an interval that captured approximately 95% of the galaxy velocities in the cluster. Explain why that interval is inappropriate for the inference you are asked to make here.b. Construct a 95% confidence interval for the mean light velocity emitted form all galaxies in cluster A2142. Interpret the result.Answers:b. Answer in 7.22 (mistake)7.22 (2 points) Attention time given to twins. Psychologists have found that twins, in their early years, tend to have lower IQs and pick up language more slowly thannontwins. (Wisconsin Twin Research Newsletter, Winter 2004). The slower intellectual growth of most twins may be caused by benign parental neglect. Suppose it is desired to estimate the mean attention time given to twins per week by their parents. A sample of 50 sets a 2.5-year-old twin boys is taken, and at the end of1 week, the attention time given to each pair is recorded. The data (in hours) are listed in the following table:20.7 16.7 22.5 1.1 2.923.5 6.4 1.3 39.6 35.610.9 7.1 46.0 23.4 29.444.1 13.8 24.3 9.3 3.415.7 46.6 10.6 6.7 5.414.0 20.7 48.2 7.7 22.220.3 34.0 44.5 23.8 20.043.1 14.3 21.9 17.5 9.636.4 0.8 1.1 19.3 14.632.5 19.1 36.9 27.9 14.0Find a 90% confidence interval for the mean attention time given to all twin boys bytheir parents. Interpret the confidence interval.Answers:7.32 (3 points) The following sample of 16 measurements was selected form a population that is approximately normally distributed:91 80 99 110 95 106 78 121 106 100 97 82 100 83 115 104a. Construct an 80% confidence interval for the population mean.b. Construct a 95% confidence interval for the population mean and compare the width of this interval with that of part a.c. Carefully interpret each of the confidence intervals, and explain why the 80% confidence interval is narrower.Answers:7.35 (5 points) Radioactive Lichen. Refer to the Lichen Radionuclide Baseline Research project at the University of Alaska, presented in Exercise 2.34 (p.47). Recall that the researchers collected 9 lichen specimens and measured the amount (in microcuries per milliliter) of the radioactive element cesum-137 for each. (The natural logarithms of the data values are saved in the Lichen file.) A MINITAB printout with summary statistics for the actual data is shown above.a. Give a point estimate for the mean amount of cesium in lichen specimens collected in Alaska.b. Give the t-value used in a small-sample 95% confidence interval for the true mean amount of cesium in Alaskan lichen specimens.c. Use the result you obtained in part b and the values of € x and sshown on the MINITAB printout to form a 95% confidence interval for the true mean amount of cesium in Alaskan lichen specimens.Answers: (a) .009 (b) 2.306 (c) .009 +/- .00377.40 (4 points) Minimizing tractor skidding distance. In planning for a new forest roadto be used for tree harvesting, planners must select the location that will minimize tractor skidding distance. In the Journal of Forest Engineering (July 1999), researchers wanted to estimate the true mean skidding distance along new road in a European forest. The skidding distance (in meters) were measured at 20 randomly selected road sites. These values are given in the accompanying table.a. Estimate, with a 95% confidence interval, the true mean skidding distance of theroad.b. Giave a practical interpretation of the interval you found in part a.c. What conditions are required for the inference you made in part b to be valid? Are these conditions reasonably satisfied?d. A logger working on the road claims that the mean skidding distance is at least 425 meters. Do you agree?Skidding488 350 457 199 285 409 435 574 439 546 385295 184 261 273 400 311 312 141 425Answers:7.50 (2 points) A random sample of 50 consumers taste-tested a new snack food. Their responses were coded (0; don not like; 1; like; 2;