A random sample of 20 houses selected from a city showed that the mean size of t

Question

A random sample of 20 houses selected from a city showed that the mean size of these houses is 1880 square feet with a standard deviation of 320 square feet. Assume that the sizes of all houses in this city have an approximate normal distribution. The upper bound the 90% confidence interval for the mean size of all houses in this city is: A. 2110 B. 1941 C. 1974 D. 1968 E. 1894 Based on a recent survey of 30 private colleges the endowment funds of all US colleges were estimated with the following 95% confidence interval: 180.975 plusminus 135.850 (in millions $) The above estimate is too wide to be of any practical importance. How should the survey design be changed in order to produce more accurate (same as precise, narrower, smaller margin of error) interval estimate? A. Decrease sample size to n = 28 B. Increase confidence level to 99% C. Increase sample size to n = 255 D. Keep everything the same, but construct an interval around the population mean. Consider H_0: mu = 12.6 H_a: mu > 12.6 with x bar = 13.75, sigma = 3.9, n = 100, alpha = 0.05? What is the p-value for this test? A. 0.002 B. 2.94 C. 0.0016 D. -0.0016 E. -2.94 It is claimed that the mean of X is greater than 910. The population standard deviation is known to be 120. The sample mean x bar = 980, and the sample size n = 35. Calculate the test statistic that should be used in one sample test for mean. A. 0.05 B. 3.45 C. 2.33 D. 0.345 E. -2.33

Explanation / Answer

12.upper bound , x+z*s/sq.root n

=1850+1.645*320/sq.root20

=1967.70

round to 1968

answer is option D

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