No.PZ2015120604000130 (选择题)

NO.PZ2015120604000130

问题如下：

We want to use z-statistic to construct confidence interval for a normally distribution. Assume the sample size is 100, sample mean is 15% and the standard deviation of sample is 25%. The significance level is supposed to be 10% , the confidence interval is :

选项：

- 10.88% to 19.13%.

10.88% to 19.13%.

10.88% to 20.57%.

解释：

B is correct.

In this case, we can know that the Confidence Interval=[Point Estimate +/- (reliability factor)*Standard error]=

$\bar{x} \pm z_{α / 2} \frac{σ}{\sqrt{n}} = 15 % \pm 1.65 \times \frac{25 %}{\sqrt{100}} = [10.88 %, 19.13 %]$

这里的Standard error 不应该是25%/ √100 吗？解析里的式子是什么意思

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星星_品职助教 · 2022年10月10日

同学你好，

standard error是25%/ √100

计息中的式子是根据 Confidence Interval=[Point Estimate +/- (reliability factor)*Standard error]代入数字后的式子，即：15%±1.65×25%/ √10

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NO.PZ2015120604000130问题如下We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis :A.- 10.88% to 19.13%.B.10.88% to 19.13%.C.10.88% to 20.57%.B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13%老师，既然题目说是正态分布，而且是选择题。想问下能否利用正态分布的特性，代入解题？正太分布中轴是均值，题干给出的置信区间应该也是轴对称。直接带入，分位点-均值19.13-15=4.13均值-分位点到均值的距离15-4.13=10.87然后选出最符合的的

2022-09-27 15:46 1 · 回答

No.PZ2015120604000130 (选择题)

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