online Exam 4 10 questions – Chapters 10, 11, 13 and hers is the copy below of hte previous exam to have an idea

I need 95% or more on it. This test has a time limit of 1 hour and 55 minutes  and one submition..:

___________

The population is the five employees in an physician’s office. The number of possible samples of 2 that could be selected from this population is

10

10

Question 2   0 out of 6 points

Suppose we have a negatively skewed population. According to the Central Limit Theorem, the distribution of a sample mean of 30 observations will

also be negatively skewed.

approach a normal distribution.

Question 3   6 out of 12 points

The foreman of the canning division of the Planters Peanuts Company observed that the amount of cashews in a 48-ounce can varies from can to can. Records indicate that the process follows the normal probability distribution with a mean of 48.5 ounces and a standard deviation of 0.25 ounces. The foreman randomly selects 16 cans from the canning line and determines that the mean amount of cashews per can is 48.6 ounces. Compute the probability that the sample of 16 cans would have a mean greater than or equal to 48.45 ounces.(CH 8)

mean= 48.5 Sd=0.25 n= 16 x= 48.6 P (z>= 48.6 ) = 48.6 -48.5 / 0.25/ sqrt 16 = 0.1 /0.0625 = 1.6  probabilty is 0.9452 94.52%

z = 48.45 – 48.5 / .25 / square root of 16 = -.05 / .0625 = -0.8 The probability that z is greater than -0.8 is .5000 + .2881 = .7881

Question 4   10 out of 12 points

RoadRunner Sports knows that the mean time it takes to process an order after it has been placed on its website is 23 minutes with a standard deviation of 8 minutes. If a random sample of 36 online orders is taken, what is the probability that the mean of the sample is less than 20 minutes?

z = 20 – 23 / 8/ sqrt 36 = -3 /1.33 = 2.2556 from table the probabiliy is 0.9759

so the proabiliy is 97.59%

z = Xbar- mu / std dev / sq rt n =

20 – 23 / 8 / sq rt 36 = – 2.25

z value for -2.25 is .4878

The likelihood of finding a z value less than -2.25 is found by (.5000 – .4878) = .0122. There is about a one percent chance that we could select a sampole of 36 online orders and find the mean processing time of the sample is 20 minutes or less, when the population mean s 23 minutes.

Question 5   8 out of 12 points

The Internal Revenue Service is studying contributions to charity. A random sample of 36 returns is selected. The population mean contribution is \$150 and the population standard deviation of  is \$20.       a. Construct a 98 percent confidence interval for the population mean.       b. Compute the 98 percent confidence interval if the population consists of 200 tax returns. (Finite population correction factor)

mean= 150 SD = 20 n= 36 t= 2.424 150 +/- 2.424 x 20/ sqrt 36 =  150-8.08 and 150=8.08 141.92 , 158.08

b) 150 +/- 2.05 x 20/ sqrt 200 = 150-2.90 and 150=2.90 147.10 152.90

a. The z value is 2.33, found by Appendix D and locating the value of .4901 in the body of the table, and reading the corresponding row and column values.   \$150 + 2.33 (\$20/square root of 36) = \$150 + \$7.77 = \$142.23 to \$157.77

b. \$150 + 2.33 (\$20/ sq root 36) (sq root 200 – 36 / 200 – 1) = \$150 + \$7.05 = \$142.95 to \$157.05

Question 6   12 out of 16 points

A manufacturer of batteries for “kids’ toys” wishes to investigate the length of time a battery will last. Test results on a sample of 10 batteries indicated a sample mean of 5.67 and a sample standard deviation of 0.57.  (SHOW ALL CALCULATIONS.)      a.  Determine the mean and the standard deviation.       b. What is the population mean? What is the best estimate of that value?       c.  Construct a 95 percent confidence interval for the population mean?       d.  Is it reasonable for the manufacturer to claim that the batteries will last 6.0 hours? (Show your calculations.)

a) mean= 5.67 SD = 0.57 b) population mean will be sample mean if the sample is normally distributed = 5.67 sample mean is the best estimate of population mean c) z for 95%=1.645 5.67 –  1.645 x 0.57/ sq rt 10 and  5.67 +  1.645 x 0.57/ sq rt 10 5.67 – 0.2967 and 5.67 + 0.2967        5.37       and 5.97  d) No the manufacturer cannot claim that batteries will last 6.0 hrs as the amximum value is 5.97 hrs.

a. The sample mean is 5.67. The standard deviation is .57 b. The population mean is not known. The best estimate is the sample mean of 5.67. c. Formula 9-2   5.67 + 2.262 (.57/sq root of 10) = 5.67 + 0.408 = 5.262 and 6.078 d. It is reasonable for the manufacturer to claim that the batteries will last 6.0 hours since 6.0 is in the interval.

Question 7   10 out of 12 points

A random sample of 100 light bulbs is selected. Sixty were found to burn for more than 1,000 hours. Develop a 90 percent confidence interval for the proportion of bulbs that will burn more than 1,000 hours.

n=100 p= 60/100 = 0.6 CI for 90% = 1.28  CI = p+/- z sqrt ( p x (1-p)/n) = 0.6 – 1.28 sqrt ( 0.6 x 0.4/100) and   0.6 + 1.28 sqrt ( 0.6 x 0.4/100) 0.6 – 0.06 and 0.6 =0.06 CI = (0.54 , 0.66)

The z value is 1.65, found by App D and locating the value of .4500 in the body of the table, and reading the corresponding row and column values.  0.60 + 1.65 (square root (0.60)(1-.60) / 100 = .60 + 0.08 = .52 and .68

Question 8   8 out of 12 points

A health maintenance organization (HMO) wants to estimate the mean length of a hospital stay. How large a sample of patient records is necessary if the HMO wants to be 99 percent confidence of the estimate and wants the estimate to be within plus or minus 0.2 days? An earlier study showed the standard deviation of the length of stay to be 0.25 days.(Round!) SHOW ALL CALCULATIONS!

n = ( z x Sd/ E)^2 = ( 2.35 x 0.25/ 0.2)^2  =8.6 or 9

n = [(2.58)(.25) / .20] squared  = 10.4 = 11

Question 9   6 out of 12 points

A large bank believes that one-third of its checking customers have used at least one of the bank’s other services during the past year. How large a sample is required to estimate the actual proportion with a range of plus or minus 0.04? Use the 98 percent level of confidence.