In a study of 420,148 cell phone​ users, 136 subjects developed cancer of the brain or nervous system. Test the claim of a somewhat common belief that such cancers are affected by cell phone use. That​ is, test the claim that cell phone users develop cancer of the brain or nervous system at a rate that is different from the rate of​ 0.0340% for people who do not use cell phones. Because this issue has such great​ importance, use a 0.005significance level. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method and the normal distribution as an approximation to the binomial distribution.       1. What is the test​ statistic?   z= _____ ​(Round to two decimal places as​ needed.)     2. What is the​ P-value?   ​P-value= _____ ​(Round to four decimal places as​ needed.)     3. What is the conclusion on the null​ hypothesis?   A. Fail to reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, α.   B. Reject the null hypothesis because the​ P-value is greater than the significance​ level, α.   C. Reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, α.   D. Fail to reject the null hypothesis because the​ P-value is greater than the significance​ level, α.   4. What is the final​ conclusion?   A. There is not sufficient evidence to support the claim that cell phone users develop cancer of the brain or nervous system at a rate that is different from the rate of​ 0.0340%.   B. There is sufficient evidence to support the claim that cell phone users develop cancer of the brain or nervous system at a rate that is different from the rate of​ 0.0340%.   C. There is sufficient evidence to warrant rejection of the claim that cell phone users develop cancer of the brain or nervous system at a rate that is different from the rate of​ 0.0340%.   D. There is not sufficient evidence to warrant rejection of the claim that cell phone users develop cancer of the brain or nervous system at a rate that is different from the rate of​ 0.0340%.

Consider the following measurements of blood hemoglobin concentrations (in g/dL) from three human populations at different geographic locations: population1 = [ 14.7 , 15.22, 15.28, 16.58, 15.10 ] population2 = [ 15.66, 15.91, 14.41, 14.73, 15.09] population3 = [ 17.12, 16.42, 16.43, 17.33] Perform ANOVA to check if any of these populations have different mean hemoglobin concentrations.  (Assume that all the ANOVA requirements such as normality, equal variances and random samples are met.)  After you perform ANOVA perform a Tukey-Kramer post-hoc test at a significance level of 0.05 to see which populations actually have different means.  As usual, round all answers to two digits after the decimal point.  (Make sure you round off to at least three digits any intermediate results in order to obtain the required precision of the final answers.) For any questions, which ask about differences in means or test statistics, which depend on differences in means provide absolute values.  In…

The effectiveness of surgery for weight loss reported here found that "The surgery was associated with significantly greater weight loss [than the control group who dieted] through 2 years (61.3 versus 11.2 pounds, P<0.001)." What test could have been used and how would it have been computed?

The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,95110,951 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 184184 of their customers and calculates that these customers used an average of 11,162kWh11,162⁢kWh of electricity last year. Assuming that the population standard deviation is 1263kWh1263⁢kWh, is there sufficient evidence to support the power company's claim at the 0.050.05 level of significance?   Step 2 of 3 :   Compute the value of the test statistic. Round your answer to two decimal places.