Question

Use the given information to find the minimum sample size required to estimate an unknown population mean . Margin of error: $121, confidence level: 99%, o = $525 O A. 125 OB. 51 O c. 72 OD. 64

Answer 1

E = margin of error = 121

= 525

The formula for sample size

n n = 29 *o 2 E

29 = Z critical value

C = confidence level = 0.99 ( 99% is converted into decimal )

a = 1- C=1-0.99 0.01

Area = 1- = = 1 0.01 =1-0.005 = 0.9950

We can find the Z score for area 0.9950 that will be Z critical value

We use a Z table

TABLE 5(a) continued Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .5000 .5040 5080 5120 51605199 .5239 .5279.5319.5359 0.1 .5398 5438 5478 .5517 5557 5596 .5636 .5675.5714 5753 0.2 .5793 .58325871 .5910 .5948 5987 .6026 .6064.6103.6141 0.3 .6179 .6217 .6255 .6293 6331.6368 .6406 .6443.6480.6517 0.4 .6554 .6591 .6628 .6664.6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 7054 7088 .7123 .7157 | .7190.7224 0.6 .7257 7291 7324 7357 .7389.7422 .7454 .7486 | .7517 .7549 0.7 .7580 .7611 7642.7673.7704 7734 .7764 .7794.7823.7852 0.8 .7881 .7910 7939.7967 7995.8023 .8051 .8078.8106 .8133 0.9 .8159 .8186 8212.8238 .8264 8289 .8315 .8340.8365 .8389 1.0 .8413 .8438 8461.8485.8508.8531 .8554 .8577.8599.8621 1.1 .8643 .8665 .8686 8708 8729.8749 .8770 .8790.8810 8830 1.2 .8849 .8869.8888 .8907 8925.8944 .8962 .8980.8997 9015 1.3 .9032 .90499066 9082 90999115 .9131 .9147.9162 9177 1.4 .9192 .9207 9222 9236 9251 9265 .9279 .9292.9306 9319 1.5 .9332 .9345 9357 9370 93829394 .9406 .9418.94299441 1.6 .9452 .9463 9474 9484 9495 9505 .9515 .9525.9535 9545 1.7 .9554 .9564 9573 9582 9591 9599 .9608 .9616.9625 9633 1.8 .9641 .9649.9656 9664 9671 9678 .9686 .9693.96999706 1.9 .9713 .9719 9726 9732 9738 9744 .9750 .9756.9761 9767 .9772 .9778 9783 9788 9793 9798 .9803 .9808.9812 9817 2.1 .9821 .9826 9830 9834 9838 9842 .9846 1.9850.9854 9857 2.2 .9861 .9864 9868 9871 9875 9878 9881 .9884.9887 9890 2.3 .9893 .9896 9898 9901 9904 9906 9909 .9911.9913 9916 2.4 .9918 99209922 99259927 9929 .9931 1.9932.9934 9936 2.5 .9938 9940 9941 9943 9945 9946 9948 .9949|| 9951 9952 2.6 .9953 99559956 9957 9959 9960 .9961 99629963 9964 2.7 .9965 9966 9967 9968 99699970 .9971 99729973 9974 2.8 9974 9975 9976 9977 9977 9978 .9979 .99799980 9981 2.9 9981 998299829983 9984 9984 .9985 .9985 9986 9986 3.0 9987 9987 9987 9988 9988 9989 .998999899990 9990 3.1 .9990 9991 9991.9991 9992.9992 .99929992 9993 9993 3.2 .9993 .9993 9994 9994 9994 9994 9994 9995 9995 9995 3.3 .9995 99959995 9996 9996 9996 .9996 .9996 9996 9997 3.4 .9997 99979997 9997 9997 9997 .9997 .9997 9997 9998 For z values greater than 3.49, use 1.000 to approximate the area. 2.0

We get area 0.9949 is less than and very close to 0.9950

Z score for area 0.9949 = Row headed number for area 0.9949 + Column headed number for area 0.9949

Z score for area 0.9949 = 2.5 + 0.07 = 2.57

Z score for area 0.9949 = Z score for area 0.9950 = 2.57

Z critical value = 2.57

= 525

E = margin of error = 121

n n = 29 *o 2 E

7 = [2.57 * 525 121

n (1349.251 121

= [11.15082645]

n = 124.3409304

We round the sample size (n ) value to the next whole number

Sample size (n ) = 125

Final answer:-

The correct choice is (A)

Sample size (n ) = 125