For the specified margin of error and confidence level, obtain a sample size that will ensure a margin of error of at most the one specified. margin of error=0.04; confidence level=90%
Q: With 99% confidence maximum air is 0.06 and the proportion is equal to 0.5. find the sample size
A: Solution
Q: For a confidence level of 90% with a sample size of 22, find the critical t value.
A: Given that, 90% confidence level Sample size = 22
Q: For a confidence level of 99% with a sample size of 14, find the critical t value.
A:
Q: For a confidence level of 80% with a sample size of 14, find the critical t value.
A: Using t table
Q: For a confidence level of 80% with a sample size of 22, find the critical t value.
A: Given information- Sample size, n = 22 Confidence level, c = 0.80 So, significance level (α) = 0.2…
Q: For a confidence level of 99% with a sample size of 29, find the critical t value.
A: Given, Sample size (n) = 29 Significance level (α)= 99% degrees of freedome = n - 1= 29 - 1 = 28
Q: For a confidence level of 99% with a sample size of 31, find the critical t value.
A: For a confidence level of 99% with a sample size of 31, find the critical t value Solution
Q: For a confidence level of 98% with a sample size of 16, find the critical t value.
A: It is given that the sample size is 16 and the level of significance is 0.02.
Q: c) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.10…
A:
Q: Find the critical value χ 2/L corresponding to a sample size of 24 and a confidence level of 95…
A:
Q: For a confidence level of 99% with a sample size of 11, find the critical t value.
A: Givenconfidence level=99%sample size(n)=11
Q: For a confidence level of 99% with a sample size of 15, find the critical t value.
A: The sample size is 15. The degrees of freedom is, df=n-1=15-1=14 The confidence level is 99% then…
Q: For a confidence level of 95% with a sample size of 16, find the critical t value.
A: We have given that, Sample size (n) = 16 and confidence level = 95% Then, We will find the…
Q: For a confidence level of 95% with a sample size of 17, find the critical t value.
A: Solution-: We have c=0.95,n=17 We find critical t-value=?
Q: For a confidence level of 90% with a sample size of 10, find the critical t value.
A: n = 10 Degrees of freedom = df = n - 1 = 10 - 1 = 9 At 90% confidence level the t is ,
Q: use the given data to find the minimum sample size required to estimate the population proportion.…
A:
Q: For a confidence level of 80% with a sample size of 28, find the critical t value.
A:
Q: Using a confidence level of 95% and a population standard deviation of 24, what is the minimum…
A: Given : Margin of error, ME = 10.0 population standard deviation, σ = 24.0…
Q: Use the given data to find the minimum sample size required to estimate the population proportion.…
A: The confidence level is 90%. Therefore, α = 0.10 and α/2=0.05. The margin of error is 0.09 Since the…
Q: For a confidence level of 98% with a sample size of 31, find the critical t value.
A: According to the provided information, α = 0.02 df = n-1 = 31-1 = 30
Q: For a confidence level of 80% with a sample size of 8, find the critical t value.
A:
Q: What sample size do they need to use if they want to estimate the proportion of branches with ATM’s…
A: The provided information are: Confidence level = 95% = 0.95 So, significance level (alpha) = 1-0.95…
Q: Use the given data to find the minimum sample size required to estimate the population proportion.…
A: Given Information: Margin of error: 0.03 confidence level: 99% p is estimated by 0.13
Q: For a confidence level of 99% with a sample size of 26, find the critical t value.
A: Given: Confidence level = 0.99 sample size (n) = 26 level of significance (α) = 1 - confidence level…
Q: For a confidence level of 90% with a sample size of 18, find the critical t value.
A: For level of significance α and sample of size n, the critical value is tα/2,n – 1 with degrees of…
Q: For a confidence level of 90% with a sample size of 31, find the critical t value.
A: Refer to the t-distribution table and look for t value corresponding to degrees of freedom=30 &…
Q: For a confidence level of 98% with a sample size of 12, find the critical value.
A:
Q: For a confidence level of 98% with a sample size of 27, find the critical t value.
A: It is given that the sample size is 27 and the level of significance is 0.02.
Q: For a confidence level of 80% with a sample size of 26, find the critical t value.
A: For a confidence level of 80% with a sample size of 26. find the critical t value ?
Q: Assume that a sample is used to estimate a population mean u. Find the margin of error M.E. that…
A: Given information: The sample mean is x-bar = 36.1. The sample standard deviation is s = 15.2. The…
Q: Using the Chi-Square x^2 distribution, for a 95%confidence level and df = 21, find X 2/L
A:
Q: For a confidence level of 90% with a sample size of 19, find the critical t value.
A: The objective is to find critical value for a confidence level of 90% with a sample size of 19.…
Q: For a confidence level of 80% with a sample size of 11, find the critical t value.
A: Given that Sample size n =11 Level of significance =alpha =0.20
Q: For a confidence level of 98% with a sample size of 17, find the critical t value.
A: Given: Sample size = n = 17, Significance level (alpha) = 1 - 0.98 = 0.02.
Q: Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.20 for…
A: Standard Deviation = σ = 1.89 Margin of Error = E = 1.2 Confidence Level = 95 Significance…
Q: For a confidence level of 98% with a sample size of 30, find the critical t value.
A:
Q: Use the given data to find the minimum sample size required to estimaate the population proportion.…
A: The confidence level is 99% then the level of significance is 0.01.
Q: For a confidence level of 95% with a sample size of 12, find the critical t value.
A: The critical t value is calculated by using the following steps:Under df column of the t…
Q: For a confidence level of 90% with a sample size of 13, find the critical t value.
A: Given ,sampl size(n)=13degrees of fredom (df)=n-1=13-1=12α=1-0.90=0.1α2=0.05
Q: Find the critical value χ 2/R corresponding to a sample size of 6 and a confidence level of 95…
A: Given,sample size(n)=6degrees of freedom(df)=n-1=6-1=5α=1-0.95=0.05α2=0.0251-α2=1-0.025=0.975
Q: Use the given data to find the minimum sample size required to estimate the population proportion.…
A: From the given information, Margin error = 0.04 Sample proportion = 0.60 Critical value: Using…
Q: For a confidence level of 99% with a sample size of 25, find the critical t value.
A: We have to find crirtical value for given data..
Q: For a confidence level of 98% with a sample size of 25, find the critical t value.
A:
Q: For a confidence level of 95% with a sample size of 28, find the critical t value.
A: where n=sample size
Q: Determine the sample size needed to estimate a population with a mean of 10 units given that the…
A: The sample size n is given as n=Zα2σE2 The values of α=0.10E=10σ=50
Q: use the given data to find the minimum sample size required to estimate the population proportion.…
A: find the minimum sample size required to estimate the population proportion. margin of error:0.07;…
Q: At 90% confidence, how large a sample should be taken to obtain a margin of error of 0.047 for the…
A: Obtain the sample size of a no preliminary estimate for p. The formula of sample size is, Use EXCEL…
Q: For the specified margin of error and confidence level, obtain a sample size that will ensure a…
A: From the provided information, Margin of error (E) = 0.05 Confidence level = 95%
Q: For a confidence level of 95% with a sample size of 15, find the critical t value
A: The degrees of freedom is,
Q: Find the critical value X 2/R corresponding to a sample size of 6 and confidence level of 95 percent
A: It is given that the sample size is 6 and level of significance is 0.05.
confidence level=90%
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- worm infestation, commonly found in children, can be treated with a drug that is effective in 92% of cases. Suppose that three children with pinworm infestation are en the drug. Complete parts (a) through (g) below. sfs 0.068 sff 0.006 fss 0.068 fsf 0.006 ffs 0.006 0.001 fff (Round to three decimal places as needed.) c. Draw a tree diagram for this problem. (Let s be a success and f be a failure.) Choose the correct tree diagram below. A. B. SSS ssf f sfs sff fss fsf ffs fff d. List the outcomes in which exactly two of the three children are cured. (Let s be a success and f be a failure.) (Use a comma to separate answers as needed.) View an exar Help me solve this y Policy | Copyright CorreSuppose the average waiting time for a customer's phone call be answered tby a company representative is three minutes. What is the median waiting tine? (Assume that the wait time has an exponential distribution, Round your answer to one decimal place.) minStandard Normal Dist. Table 1 z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09-3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002-3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003-3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005-3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007-3.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010-2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014-2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019-2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026-2.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036-2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048-2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064-2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084-2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110-2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143-2.0 .0228 .0222 .0217 .0212…
- Standard Normal Dist. Table 1 z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09-3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002-3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003-3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005-3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007-3.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010-2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014-2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019-2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026-2.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036-2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048-2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064-2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084-2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110-2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143-2.0 .0228 .0222 .0217 .0212…Standard Normal Dist. Table 1 z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09-3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002-3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003-3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005-3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007-3.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010-2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014-2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019-2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026-2.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036-2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048-2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064-2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084-2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110-2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143-2.0 .0228 .0222 .0217 .0212…Age (YRS) Mass (KG) Cigarettes (Per Day) Risk Index (R.I.) (--) 46 113 40 974.5 35 96 32 670.68 43 81 26 409.26 39 93 50 1484.5 22 66 25 500 58 55 33 631.87 60 72 15 350.75 54 93 19 347.09 32 55 12 308.78 63 115 24 580.74 30 105 20 320.5 24 90 19 381.59 55 65 15 274.25 30 115 6 322.66 40 115 21 411.11 34 80 22 396.48 31 88 22 906 70 57 33 616.87 35 116 43 1072.07 57 101 31 589.41 44 68 36 718.56 66 118 33 790.37 46 120 44 1129.84 20 72 19 487 64 63 35 665.25 30 113 3 204.26 32 70 20 312 37 68 23 299.67 52 88 26 418.76 64 70 46 1299.36 23 87 23 422.17 42 53 50 1500.5 54 64 15 299.75 35 95 49 1536.99 51 95 39 912.69 61 105 20…
- Assuming that mean annual temperature uniformly distributed between 100C and 180C, calculate the average annual mean temperature.Price_($000)/Sq Ft 1/Sq Ft Location 14.0542 0.0653 0 17.2907 0.0524 0 18.3029 0.00649 0 15.7341 0.04374 0 14.8521 0.01913 0 15.3572 0.0197 0 18.3278 0.01583 0 15.8483 0.41563 0 17.392 0.29976 0 15.8826 0.1798 0 15.5492 0.00816 0 14.619 0.0325 0 15.7633 0.21829 0 15.1289 0.015939 0 17.2874 0.18765 0 15.6938 0.00986 0 15.3672 0.03096 0 15.3126 0.01917 0 17.4008 0.06055 0 16.1066 0.0611 0 16.6637 0.1171 1 16.5504 0.05552 1 17.0575 0.00657 1 17.9466 0.1407 1 17.8913 0.0648 1 15.5396 0.03339 1 18.6629 0.03443 1 18.0774 0.019 1 20.8122 0.20162 1 18.3237 0.319 1Find the value of z if the area under a standard normal curve (a) to the right of z is 0.3228; (b) to the left of z is 0.1271; (c) between 0 and z, with z> 0, is 0.4890; and (d) between -zand z, with z> 0, is 0.9500. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table.
- determine if the frequency of deficiencies of variable (medication errors) are statistically significantly different between for-profit and non-profit nursing homes, interpret the results for the chi-square?1/60 why is the answer on calculator (0.0166666667)determine if the frequency of deficiencies of variable (significant medications) are statistically significantly different between for-profit and non-profit nursing homes, interpret the results for the chi-square?