70f10 -/10 ;= Part 3 (c) A 92 % condence interval for a standard deviation. Round your answer to two decimal places. :z*= i 7 of 10 - / 10 E . . . Part 2 (b) An 82% confidence interval for a slope. Round your answer to two decimal places.9 of 10 - /10 WIN - = . . . Part 2 (b) Is it plausible that there is no difference between men and women in how likely they are to use online dating? Use the confidence interval from part (a) to answer. O Yes O NoPart 2 (b) 2 = 4.12for a right tail test for a proportion Round your answer to three decimal places. pvalue = 6 of 10 - 18 . . . View Policies Current Attempt in Progress MORE BENEFITS OF EATING ORGANIC Using specific data, we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. This exercise asks you to conduct a hypothesis test using additional data. In this case, we are testing Ho : Po = Pc Ha : Po > Pc where po and pc represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Use a 5% significance level. ect of Organic Potatoes After 11 Days10f10 > -/4 ;= View Policies Current Attempt in Progress The following is a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Test H0 2}) = 0.29 vs Ha :p I \\ _'n O NI o View Policies Current Attempt in Progress More Information From the Online Dating Survey A survey conducted in July 2015 asked a random sample of American adults whether they had ever used online dating (either an online dating site or a dating app on their cell phone) Comparing Males to Females In the survey, 17 % of the men said they had used online dating, while 14 % of the women said they had. Part 1 0a) Find a 99 % condence interval for the difference in the proportion saying they 2 of 10 - 14 . . . View Policies Current Attempt in Progress The following is a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Test Ho : M = 8 vs Ha : M # 8 when the sample has n = 80, x = 11.5, and s = 0.90 with SE = 0.10. Find the value of the standardized z-test statistic. Round your answer to two decimal places.Part 3 (c) z = -1.60for a two-tailed test for a difference in means Round your answer to three decimal places. p-value =60f10 -/8 ;= Effect of Organic Potatoes After 1 1 Days After 11 days, the proportion of fruit ies eating organic potatoes still alive is 0.65 , while the proportion still alive eating conventional potatoes is 0.61 . The standard error for the difference in proportions is 0.033. What is the value of the test statistic? Round your answer to two decimal places. 2: What is the p-value? Round your answer to three decimal places. p-value = What is the conclusion? C5 0 40f10 -/10 25 View Policies Current Attempt in Progress Find the p-value based on a standard normal distribution for each of the following standardized test statistics. Part 1 (a) z = -1.13 fora left tailtest fora mean Round your answer to three decimal places. p-value = 10 of 10 Pc where po and pc represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Use a 5% significance level. ect of Organic Soybeans After 5 Days5 of 10 - 18 E . . . ROUTIu your allSVVCI TO LI CC UCCITIal Plates. p-value = What is the conclusion? C Reject H_O. Do not reject H_O.30f10 -/4 = View Policies Current Attempt in Progress Find the area in the right tail more extreme than 2 = 0.85 in a standard normal distribution. Round your answer to three decimal places. Area = 8 of 10 - 18 . . . WN - E View Policies Current Attempt in Progress Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A 90 % confidence interval for a mean u if the sample has n = 40 with x = 22.7 and s = 5.6, and the standard error is SE = 0.89. Round your answers to three decimal places. The 90 % confidence interval is HI to10 of 10 - 18 E . . . American adults ages 53 to 64 to use online dating, the standard error is 0.016. Find a 99% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating. Round your answers to three decimal places. The 99% confidence interval is HI to