开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)|3.2.5(安卓)APP下载

March18龚献 · 2022年02月07日

25000/6这步,6是哪个条件出来的?

NO.PZ2018062016000132

问题如下:

Claire believes that the mean price of the commercial residential building is greater than $155,000. The population standard deviation is $25,000. A random sample of 36 commercial residential buildings in the area has a mean price of $159,750. Claire conduct the hypothesis test at a 1% level of significance. The value of the calculated test statistic is closet to:

选项:

A.

1.14.

B.

0.19.

C.

1.33.

解释:

A is correct. This is a one-tailed hypothesis since the alternative hypothesis only allows values greater than the hypothesized value. z=159750-155000/(25000/6)=1.14

25000/6这步,6是哪个条件出来的?
1 个答案

星星_品职助教 · 2022年02月07日

同学你好,

根据中心极限定理,样本均值的标准差(标准误)为总体标准差σ除以根号n。

由此根据“The population standard deviation is $25,000. A random sample of 36”,得到样本均值( the mean price of the commercial residential building)的标准差即标准误=25,000 / 根号36=25,000 / 6

  • 1

    回答
  • 0

    关注
  • 364

    浏览
相关问题

NO.PZ2018062016000132 问题如下 Claire believes ththe mepriof the commerciresintibuilng is greater th$155,000. The population stanrviation is $25,000. A ranm sample of 36 commerciresintibuilngs in the area ha mepriof $159,750. Claire conthe hypothesis test a 1% level of significance. The value of the calculatetest statistic is closet to: A.1.14. B.0.19. C.1.33. A is correct. This is a one-tailehypothesis sinthe alternative hypothesis only allows values greater ththe hypothesizevalue. z=159750-155000/(25000/6)=1.14 这个25000不是总体样本方差嘛?population不是总体方差嘛?请老师解惑下。

2023-04-27 15:25 1 · 回答

NO.PZ2018062016000132 问题如下 Claire believes ththe mepriof the commerciresintibuilng is greater th$155,000. The population stanrviation is $25,000. A ranm sample of 36 commerciresintibuilngs in the area ha mepriof $159,750. Claire conthe hypothesis test a 1% level of significance. The value of the calculatetest statistic is closet to: A.1.14. B.0.19. C.1.33. A is correct. This is a one-tailehypothesis sinthe alternative hypothesis only allows values greater ththe hypothesizevalue. z=159750-155000/(25000/6)=1.14 这个知识点对应哪一章节

2022-06-29 19:26 1 · 回答

NO.PZ2018062016000132 0.19. 1.33. A is correct. This is a one-tailehypothesis sinthe alternative hypothesis only allows values greater ththe hypothesizevalue. z=159750-155000/(25000/6)=1.14求单尾假设significant level是1%还是0.5%

2021-11-09 12:09 1 · 回答

NO.PZ2018062016000132 我用上方公式算对了这题。但我不明白分子为什么不可以是the hypothesizevalue减去X̄? 如果是要衡量the hypothesizevalue在分布上离均值X̄多远,既然分布是左右对称,【the hypothesizevalue减去X̄再除以标准误】和【X̄减去the hypothesizevalue再除以标准误】,算出来的数值只是相差了一个正负号。我该怎么理解公式的分子【X̄减去the hypothesizevalue】的意思?

2021-09-15 16:18 1 · 回答