12 In another region 5% of the x values from samples of size n = 25 are under $4

Question

12 In another region 5% of the x values from samples of size n = 25 are under $48,000 and 5% are over $57,000. From this sampling distribution information we can conclude that the population mean cost of a four-year college degree is = _______. µ a $50,500 b $51,500 c $52,500 d $53,500 13 In the previous question, the population standard deviation is = ______. a $12,980 b $13,720 c $14,240 d $14,760 Questions 14-17 are based on the following: The mean annual Medicare spending per enrollee is $13,800 with a standard deviation of $3,960. Answer questions 14-17 based on the sampling distribution of x for random samples of size n = 81 enrollees. 14 The fraction of sample means falling within ±$600 from the population mean is ______. a 0.9026 b 0.8849 c 0.8675 d 0.8262 15 95% of all x values from samples of size n = 81 deviate from the population mean of $13,800 by no more than ±$______. a $939 b $903 c $862 d $790 12 In another region 5% of the x values from samples of size n = 25 are under $48,000 and 5% are over $57,000. From this sampling distribution information we can conclude that the population mean cost of a four-year college degree is = _______. µ a $50,500 b $51,500 c $52,500 d $53,500 13 In the previous question, the population standard deviation is = ______. a $12,980 b $13,720 c $14,240 d $14,760 Questions 14-17 are based on the following: The mean annual Medicare spending per enrollee is $13,800 with a standard deviation of $3,960. Answer questions 14-17 based on the sampling distribution of x for random samples of size n = 81 enrollees. 14 The fraction of sample means falling within ±$600 from the population mean is ______. a 0.9026 b 0.8849 c 0.8675 d 0.8262 15 95% of all x values from samples of size n = 81 deviate from the population mean of $13,800 by no more than ±$______. a $939 b $903 c $862 d $790

Explanation / Answer

12) mean =(48000+57000)/2=52500

option C

13)here margin of error E=(57000-48000)/2=4500

and for 90% CI, z=1.64

population std deviation =(E/Z) *(n)1/2 = 13720

14) std error =std deviation/(n)1/2 =440

P(-600/440<Z<600/440)=P(-1.3636<Z<1.3636)=0.9137-0.0863=0.8273

option D

15)for 95% CI, z=1.96

hence margin of error =z*std error =1.96*440=862

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