統計學!信賴區間! 很急

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!
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)

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