Question

1) Find the critical value z Subscript c necessary to form a confidence interval at the level of confidence shown below. cequals0.85

2)Level of Confidence z Subscript c ​90% 1.645 ​95% 1.96 ​99% 2.575 Find the margin of error for the given values of​ c, sigma​, and n. cequals0.95​, sigmaequals2.7​, nequals 81

3)Use the confidence interval to find the estimated margin of error. Then find the sample mean. A biologist reports a confidence interval of left parenthesis 2.1 comma 3.7 right parenthesis when estimating the mean height​ (in centimeters) of a sample of seedlings.

4) You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 31 business​ days, the mean closing price of a certain stock was ​$121.32. Assume the population standard deviation is ​$10.56.

5)You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 70 ​dates, the mean record high daily temperature in a certain city has a mean of 85.63degreesF. Assume the population standard deviation is 14.87degreesF.

6)Determine the minimum sample size required when you want to be 99​% confident that the sample mean is within one unit of the population mean and sigmaequals11.1. Assume the population is normally distributed.

7)A cheese processing company wants to estimate the mean cholesterol content of all​ one-ounce servings of a type of cheese. The estimate must be within 0.71 milligram of the population mean. ​(a) Determine the minimum sample size required to construct a 95​% confidence interval for the population mean. Assume the population standard deviation is 3.19 milligrams. ​(b) The sample mean is 33 milligrams. Using the minimum sample size with a 95​% level of​ confidence, does it seem likely that the population mean could be within 3​% of the sample​ mean? within 0.3​% of the sample​ mean? Explain. z 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 minus3.4 0.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 minus3.3 0.0003 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0005 0.0005 0.0005 minus3.2 0.0005 0.0005 0.0005 0.0006 0.0006 0.0006 0.0006 0.0006 0.0007 0.0007 minus3.1 0.0007 0.0007 0.0008 0.0008 0.0008 0.0008 0.0009 0.0009 0.0009 0.0010 minus3.0 0.0010 0.0010 0.0011 0.0011 0.0011 0.0012 0.0012 0.0013 0.0013 0.0013 minus2.9 0.0014 0.0014 0.0015 0.0015 0.0016 0.0016 0.0017 0.0018 0.0018 0.0019 minus2.8 0.0019 0.0020 0.0021 0.0021 0.0022 0.0023 0.0023 0.0024 0.0025 0.0026 minus2.7 0.0026 0.0027 0.0028 0.0029 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 minus2.6 0.0036 0.0037 0.0038 0.0039 0.0040 0.0041 0.0043 0.0044 0.0045 0.0047 minus2.5 0.0048 0.0049 0.0051 0.0052 0.0054 0.0055 0.0057 0.0059 0.0060 0.0062 minus2.4 0.0064 0.0066 0.0068 0.0069 0.0071 0.0073 0.0075 0.0078 0.0080 0.0082 minus2.3 0.0084 0.0087 0.0089 0.0091 0.0094 0.0096 0.0099 0.0102 0.0104 0.0107 minus2.2 0.0110 0.0113 0.0116 0.0119 0.0122 0.0125 0.0129 0.0132 0.0136 0.0139 minus2.1 0.0143 0.0146 0.0150 0.0154 0.0158 0.0162 0.0166 0.0170 0.0174 0.0179 minus2.0 0.0183 0.0188 0.0192 0.0197 0.0202 0.0207 0.0212 0.0217 0.0222 0.0228 minus1.9 0.0233 0.0239 0.0244 0.0250 0.0256 0.0262 0.0268 0.0274 0.0281 0.0287 minus1.8 0.0294 0.0301 0.0307 0.0314 0.0322 0.0329 0.0336 0.0344 0.0351 0.0359 minus1.7 0.0367 0.0375 0.0384 0.0392 0.0401 0.0409 0.0418 0.0427 0.0436 0.0446 minus1.6 0.0455 0.0465 0.0475 0.0485 0.0495 0.0505 0.0516 0.0526 0.0537 0.0548 minus1.5 0.0559 0.0571 0.0582 0.0594 0.0606 0.0618 0.0630 0.0643 0.0655 0.0668 minus1.4 0.0681 0.0694 0.0708 0.0721 0.0735 0.0749 0.0764 0.0778 0.0793 0.0808 minus1.3 0.0823 0.0838 0.0853 0.0869 0.0885 0.0901 0.0918 0.0934 0.0951 0.0968 minus1.2 0.0985 0.1003 0.1020 0.1038 0.1056 0.1075 0.1093 0.1112 0.1131 0.1151 minus1.1 0.1170 0.1190 0.1210 0.1230 0.1251 0.1271 0.1292 0.1314 0.1335 0.1357 minus1.0 0.1379 0.1401 0.1423 0.1446 0.1469 0.1492 0.1515 0.1539 0.1562 0.1587 minus0.9 0.1611 0.1635 0.1660 0.1685 0.1711 0.1736 0.1762 0.1788 0.1814 0.1841 minus0.8 0.1867 0.1894 0.1922 0.1949 0.1977 0.2005 0.2033 0.2061 0.2090 0.2119 minus0.7 0.2148 0.2177 0.2206 0.2236 0.2266 0.2296 0.2327 0.2358 0.2389 0.2420 minus0.6 0.2451 0.2483 0.2514 0.2546 0.2578 0.2611 0.2643 0.2676 0.2709 0.2743 minus0.5 0.2776 0.2810 0.2843 0.2877 0.2912 0.2946 0.2981 0.3015 0.3050 0.3085 minus0.4 0.3121 0.3156 0.3192 0.3228 0.3264 0.3300 0.3336 0.3372 0.3409 0.3446 minus0.3 0.3483 0.3520 0.3557 0.3594 0.3632 0.3669 0.3707 0.3745 0.3783 0.3821 minus0.2 0.3859 0.3897 0.3936 0.3974 0.4013 0.4052 0.4090 0.4129 0.4168 0.4207 minus0.1 0.4247 0.4286 0.4325 0.4364 0.4404 0.4443 0.4483 0.4522 0.4562 0.4602 minus0.0 0.4641 0.4681 0.4721 0.4761 0.4801 0.4840 0.4880 0.4920 0.4960 0.5000 z 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00

Answer 1

1)

Critical value = +/- 1.44

2)

margin of error for 90%

z value at 90% =1.645

ME = z *(s/sqrt(n))
= 1.645 *(2.7/sqrt(81))
= 0.4935

z value at 95% =1.96

ME = z *(s/sqrt(n))
= 1.96 *(2.7/sqrt(81))
= 0.5880

z value at 99% = 2.576

ME = z *(s/sqrt(n))
= 2.576 *(2.7/sqrt(81))
= 0.7725

4)

z value at 90% = 1.645

CI = mean +/- z *(s/sqrt(n))
= 121.32 +/- 1.645 *(10.56/sqrt(31))
= (118.20,124.44)

z value at 95% = 1.96

CI = mean +/- z *(s/sqrt(n))
= 121.32 +/- 1.96 *(10.56/sqrt(31))
= (117.60,125.04)