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- Assume the given confidence interval for the population mean of the girth of Saguaro cacti in Saguaro National Park, μμ. Find the margin of error, E.54.6 inches < μμ < 60 inches E = inchesGiven that you need to construct a 90% confidence interval, having 12 measurements, and you don't know the value of σ , what would be the correct t -score to use? Find tn−1,α/2 (round to 3 d.p)Use the given data to construct a confidence interval for the population proportion p of the requested level. =x53 , =n68 , confidence level 80% Round the answers to at least three decimal places.
- A researcher estimates the 95% CI for a sample with a mean of M = 9.0 and a standard error (sM) of 1.03. What is the confidence interval at this level of confidence?A New York Times article reported that a survey conducted in 2014 included 36,000 adults, with 3.72% of them being regular users of e-cigarettes. Because e-cigarette use is relatively new, there is a need to obtain today's usage rate. How many adults must be surveyed now if a confidence level of 90% and a margin of error of 2.5 percentage points are wanted? Complete parts (a) through (c) below. a. Assume that nothing is known about the rate of e-cigarette usage among adults. n = ____________ b. Use the results from the 2014 survey. n = ______________1. The best point estimator for the average wieght of current PBA players is ? kilograms. 2. At 985 confidence, the estimate of the minimum weight of all PBA players is ? kilograms.
- For question d, if appropriate, use the two proportions z interval procedure to find the specified confidence interval.Refer to the accompanying data set and construct a 90 % confidence interval estimate of the mean pulse rate of adult females; then do the same for adult males. Compare the results. Construct a 90 % confidence interval of the mean pulse rate for adult females. nothing bpmless than muless thannothing bpm (Round to one decimal place as needed.) Construct a 90 % confidence interval of the mean pulse rate for adult males. nothing bpmless than muless thannothing bpm (Round to one decimal place as needed.) Compare the results. A. The confidence intervals do not overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males. B. The confidence intervals overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males. C. The confidence intervals overlap, so it appears that adult males have a significantly…15 The interval estimate with a 95% confidence level is, a 0.522 0.607 b 0.532 0.597 c 0.539 0.589 d 0.545 0.584
- Refer to the accompanying data set and construct a 95 % confidence interval estimate of the mean pulse rate of adult females; then do the same for adult males. Compare the results. Construct a 95 % confidence interval of the mean pulse rate for adult females. nothing bpmless than muless thannothing bpm (Round to one decimal place as needed.) Construct a 95 % confidence interval of the mean pulse rate for adult males. nothing bpmless than muless thannothing bpm (Round to one decimal place as needed.) Compare the results. A. The confidence intervals overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males. B. The confidence intervals overlap, so it appears that adult males have a significantly higher mean pulse rate than adult females. C. The confidence intervals do not overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males. D. The confidence…The measured voltage amounts from a sample of 27 households have a mean of 123.59 volts and a sample standard deviation of 1.38 volt. a) Find a 99% confidence interval to estimate the population mean of the measured voltage amounts for all households in America. b) Write a conclusion. c) Use the TI 84 method to verify the answer.What price do farmers get for their seedless watermelon crops? In a recent month, a simple random sample of 31 farming regions gave a sample mean of $12.15 per 100 pounds of seedless watermelon. Assumer that σ=$1.46 per 100 pounds. Find the sample size necessary for a 90% confidence level with a margin of error E=0.61 for the mean price per 100 pounds of seedless watermelon. (Increase to the next whole number.)