1.The average temperature during the summer months for the northeastern part of the United states is【 67.0°】.A sample of 【10】 cities had an average temperature of 【69.6°】 for the summer of 【1995】. The standard deviation of the sample is 【1.1°】. At α= 【0.10】, can it be concluded that the summer of 【1995】 was warmer than average.
2.A contractor desires to build new homes with fireplaces. He read in survey that 【80%】 of all home buyers want a fireplace. To test this claim,he selected a sample of 【30】 home buyters and found that 【20】 wanted a fireplace.Use α= 【0.02】 to test the claim.
3.A film editor feels that the standard deviation for the number of minutes in a video is 【3.4】 minutes . A sample of 【24】 videos has a standard deviation of 【4.2】 minutes . At α= 【0.05】, is the sample standard deviation different from what the editor hypothesized
4.A manufacturer claims that the standard deviation of the drying time of a certain type of paint is 【18】 minutes .A sample of five test pancels produced a standard deviation of 【21】 minutes .Test the claim at α=【0.05】.
- ?Lv 71 decade agoFavorite Answer
1. a=0.10, 信賴區間的Z值為1.645. 用Z-test, Z=(X-mo)/(s/n)=(69.6-67.0)/(1.1/10)=7.47 > 1.645, 所以1995年的夏天較熱.
2. a=0.02, 信賴區間的Z值為2.33. p=0.8, p'=20/30=0.667, n=30, 用one-proportion z-test, Z=|p-p'|/(p(1-p)/n)=(0.8-0.667)/(0.667*0.333)/30)=1.549 < 2.33, null hypothesis rejected. The claim is accepted.
3. Chi-square hypothesis is s=so, The upper critical value for a=0.05, N=24 is 38.076. T=(24-1)*(4.2/3.4)2=35.096 < 38.076. The hypothesis is accepted.
4. Chi-square hypothesis is s=so, The upper critical value for a=0.05, N=5 is 11.143. T=(5-1)*(21/18)2
2008-06-07 06:01:35 補充：
4. T=5.444<11/143. The claim is accepted.