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統計學!信賴區間! 很急 20點
4.Travel Plans You are a travel agent and wish to estimate, with 95% confidence, the proportion of vacationers who plan to travel outside the United States in the next 12 months. Your estimate must be accurate within 3% of the true proportion. NO preliminary estimate is available. Find the minimum sample size... 顯示更多 4.Travel Plans You are a travel agent and wish to estimate, with 95% confidence, the proportion of vacationers who plan to travel outside the United States in the next 12 months. Your estimate must be accurate within 3% of the true proportion. NO preliminary estimate is available. Find the minimum sample size needed. Find the minimum sample size needed, using a prior study that found that 26% of the respondents said they planned to travel outside the United States in the next 12 months. Compare the results from parts and . 4.Travel計劃您是旅行代理商，並希望估計，有95％的信心，誰的比例度假者境外旅行計劃，美國在未來 12個月。你估計要準確3％以內的真實比例。（a）沒有初步的估計。查找所需的最低樣本量。 （B）查找所需的最低樣本量，採用事先研究發現，26％的受訪者表示，他們計劃前往美國以外的國家在未來 12個月。 （C）比較的結果，從第（A）及（b）。 5.In a random sample of 15 CD players brought in for repair, the average repair cost was $80 and the standard deviation was $14. Construct a 90% confidence interval for μ. Assume the repair costs are normally distributed. 5.在隨機抽樣的15個 CD播放機帶來的維修，平均修理費用為80元，標準偏差為14美元。構建了90％置信區間為美國承擔有關的維修費用是正態分佈的。 可以幫我解答一下嗎? 很急很急!
最佳解答:
您好，舒凡很高興為您說明4.Travel Plans You are a travel agent and wish to estimate, with 95% confidence, the proportion of vacationers who plan to travel outside the United States in the next 12 months. Your estimate must be accurate within 3% of the true proportion. NO preliminary estimate is available. Find the minimum sample size needed. Find the minimum sample size needed, using a prior study that found that 26% of the respondents said they planned to travel outside the United States in the next 12 months. Compare the results from parts and . 4.Travel計劃您是旅行代理商，並希望估計，有95％的信心，誰的比例度假者境外旅行計劃，美國在未來 12個月。你估計要準確3％以內的真實比例。（a）沒有初步的估計。查找所需的最低樣本量。由於二項分配的變異數=p*(1-p)若不知道P，則使用變異數的極大值，當p=1/2 時，p*(1-p)有極大值因此用p=1/2進行估計信賴區間的容差值=Z((1-信心水準)/2)*估計的標準差=√(變異數/樣本數)= √(( p*(1-p)/n)由於p=1/2 ,因此√(( p*(1-p)/n)=1/2/√nZ((1-0.95)/2)=1.960.03=1.96*0.5/√nN=1067.11 ,取1068 （B）查找所需的最低樣本量，採用事先研究發現，26％的受訪者表示，他們計劃前往美國以外的國家在未來 12個月。估計的標準差=√(( p*(1-p)/n)= √((0.26*0.74)/n)=0.43/√n0.03=1.96*0.43/√n , n=789.23取790 （C）比較的結果，從第（A）及（b）。 若已知比例，則變異數會變小，此時，可以用比較小的樣本數，達到相同的信賴區間誤差範圍 5.In a random sample of 15 CD players brought in for repair, the average repair cost was $80 and the standard deviation was $14. Construct a 90% confidence interval for μ. Assume the repair costs are normally distributed. 5.在隨機抽樣的15個 CD播放機帶來的維修，平均修理費用為80元，標準偏差為14美元。構建了90％置信區間。假設維修費用是正態分佈的。由於維修費用是常態，因此即使樣本數很小，但仍可以Z檢定(一般小樣本需要用T值替代Z值)Z((1-0.9)/2)=1.645信賴區間=平均值±1.645*標準差/√n=80±1.645*14/√15=80±5.946=(74.054~85.946)
其他解答:D4CDE60C03A2F1BC
統計學!信賴區間! 很急 20點
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發問:4.Travel Plans You are a travel agent and wish to estimate, with 95% confidence, the proportion of vacationers who plan to travel outside the United States in the next 12 months. Your estimate must be accurate within 3% of the true proportion. NO preliminary estimate is available. Find the minimum sample size... 顯示更多 4.Travel Plans You are a travel agent and wish to estimate, with 95% confidence, the proportion of vacationers who plan to travel outside the United States in the next 12 months. Your estimate must be accurate within 3% of the true proportion. NO preliminary estimate is available. Find the minimum sample size needed. Find the minimum sample size needed, using a prior study that found that 26% of the respondents said they planned to travel outside the United States in the next 12 months. Compare the results from parts and . 4.Travel計劃您是旅行代理商，並希望估計，有95％的信心，誰的比例度假者境外旅行計劃，美國在未來 12個月。你估計要準確3％以內的真實比例。（a）沒有初步的估計。查找所需的最低樣本量。 （B）查找所需的最低樣本量，採用事先研究發現，26％的受訪者表示，他們計劃前往美國以外的國家在未來 12個月。 （C）比較的結果，從第（A）及（b）。 5.In a random sample of 15 CD players brought in for repair, the average repair cost was $80 and the standard deviation was $14. Construct a 90% confidence interval for μ. Assume the repair costs are normally distributed. 5.在隨機抽樣的15個 CD播放機帶來的維修，平均修理費用為80元，標準偏差為14美元。構建了90％置信區間為美國承擔有關的維修費用是正態分佈的。 可以幫我解答一下嗎? 很急很急!
最佳解答:
您好，舒凡很高興為您說明4.Travel Plans You are a travel agent and wish to estimate, with 95% confidence, the proportion of vacationers who plan to travel outside the United States in the next 12 months. Your estimate must be accurate within 3% of the true proportion. NO preliminary estimate is available. Find the minimum sample size needed. Find the minimum sample size needed, using a prior study that found that 26% of the respondents said they planned to travel outside the United States in the next 12 months. Compare the results from parts and . 4.Travel計劃您是旅行代理商，並希望估計，有95％的信心，誰的比例度假者境外旅行計劃，美國在未來 12個月。你估計要準確3％以內的真實比例。（a）沒有初步的估計。查找所需的最低樣本量。由於二項分配的變異數=p*(1-p)若不知道P，則使用變異數的極大值，當p=1/2 時，p*(1-p)有極大值因此用p=1/2進行估計信賴區間的容差值=Z((1-信心水準)/2)*估計的標準差=√(變異數/樣本數)= √(( p*(1-p)/n)由於p=1/2 ,因此√(( p*(1-p)/n)=1/2/√nZ((1-0.95)/2)=1.960.03=1.96*0.5/√nN=1067.11 ,取1068 （B）查找所需的最低樣本量，採用事先研究發現，26％的受訪者表示，他們計劃前往美國以外的國家在未來 12個月。估計的標準差=√(( p*(1-p)/n)= √((0.26*0.74)/n)=0.43/√n0.03=1.96*0.43/√n , n=789.23取790 （C）比較的結果，從第（A）及（b）。 若已知比例，則變異數會變小，此時，可以用比較小的樣本數，達到相同的信賴區間誤差範圍 5.In a random sample of 15 CD players brought in for repair, the average repair cost was $80 and the standard deviation was $14. Construct a 90% confidence interval for μ. Assume the repair costs are normally distributed. 5.在隨機抽樣的15個 CD播放機帶來的維修，平均修理費用為80元，標準偏差為14美元。構建了90％置信區間。假設維修費用是正態分佈的。由於維修費用是常態，因此即使樣本數很小，但仍可以Z檢定(一般小樣本需要用T值替代Z值)Z((1-0.9)/2)=1.645信賴區間=平均值±1.645*標準差/√n=80±1.645*14/√15=80±5.946=(74.054~85.946)
其他解答:D4CDE60C03A2F1BC
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