H u=4.2 versus H: u +4.2 Hor then H(Choose one be rejected at the 0.05 Since the 95%% confidence interval (Choose one) contains does not contain Will/will noe
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- We have a gantry crane that handles one TEU container in 4 minutes. A new gantry crane developer claims that their machine handles one TEU container in 3 minutes. If this is statistically significant, we will purchase this new machine. We have selected 10 random container handling operation of this machine and tested their difference from the 4 minutes by ONE SAMPLE T TEST in the 5% confidence level. We have found T STAT as -3.323. What are the H0 and Ha for this question? Should we purchase this new machine or not? Why?A 95% confidence interval of the mean is (10, 15). The result of testing Ho:H=12 against H:u#12, at 0.05 level of significance, will be Do not reject the null hypothesis because 12 is between 10 and 15 There is insufficient information to make any conclusions about the hypothesis test Reject the null hypothesis because 12 is more than 10 Do not reject the the null hypothesis because 12 is less than 15"Trydint" bubble-gum company claims that 4 out of 10 people prefer their gum to "Eklypse". Test their claim at the 95 confidence level. The null and alternative hypothesis in symbols would be: Ο Η : μ 2 0.4 Hy:μ 0.4 H₁:p 0.4 Ο Ho: μ 0.4 The null hypothesis in words would be: O The average of people that prefer Trydint gum is 0.4. The proportion of people in a sample that prefers Trydint gum is 0.4. O The proportion of all people that prefer Trydint gum is less than 0.4. O The proportion of all people that prefer Trydint gum is 0.4 O The proportion of all people that prefer Trydint gum is greater than 0.4. O The average of people that prefer Trydint gum is not 0.4. O The proportion of people in a sample that prefer Trydint gum is not 0.4 Based on a sample of 100 people, 19 said they prefer "Trydint" gum to "Eklypse". The point estimate (statistic) is: (to 3 decimals) Percentage Confidence z*-Value 80 1.28 90 1.645 95 1.96 98 2.33 99 2.58 p. (1-p) The Standard Error can be found using…
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- A survey of 260 young professionals found that one-fifth of them use their cell phones primarily for e-mail. Can you conclude statistically that the population proportion who use cell phones primarily for e-mail is less than 0 237 Use a 95% confidence interval The 95% confidence interval is. As 0.23 is population proportion is less than 0.23. (Use ascending order Round to four decimal places as needed.) of the confidence interval, we conclude that theSuppose you have a model Y = B₁ + B₂X₂ + uż that satisfies all CNLRM assumption with 8 observations, and the o2 = 36. What is the 95% confidence interval of o²? (17.17, 98.19) (18.63, 132.46) (17.75, 145.56) (14.95, 174.62) Do not have enough informationAn article claims that babies spend twice as much time in REM (rapid eye movement) sleep than adults. This implies that babies have an average of 6 hours of REM sleep per night. A doctor would like to test the hypotheses = 6 versus ≠ 6 where = the true mean amount of REM sleep per night for babies. A 95% confidence interval for the true mean amount of REM sleep per night for babies is (5.45, 6.25) hours. Based on the confidence interval, what conclusion would you make for a test of these hypotheses? The 95% confidence interval include as a plausible value, so we H0. We convincing evidence that the true mean amount of REM sleep per night for babies 6 hours.
- A random sample of ni = 270 people who live in a city were selected and 99 identified as a "dog person." A random sample of n2 = 108 people who live in a rural area were selected and 51 identified as a "dog person." Find the %3D 98% confidence interval for the difference in the proportion of people that live in a city who identify as a "dog person" and the proportion of people that live in a rural area who identify as a "dog person." Round answers to to 4 decimal places. < pi – P2 < -A survey of a random sample of 256 college students found that 36% said they frequently experience stress in their daily lives. The pollsters wish to construct a 9o% confidence interval for the true proportion of students that experience stress. 36(1-.36) A 0.36 1.28 256 36(1-36) 0.36 +1.64/ 256 36(1-36) C 0.36 +1.96 256 90(1-90) D 0.90 +1.64/ 256A scientist claims that only 67% of geese in his area fly south for the winter. He tags 70 random geese in the summer and finds that 20 of them do not fly south in the winter. If α= 0.05, is the scientist's belief warranted? Test by using confidence interval. Round confidence interval to nearest thousandth. Show step1-step3 included conclusion.