Assume that a sample is used to estimate a population mean μ. Find the margin of error M.E. that corresponds to a sample of size 14 with a mean of 25.7 and a standard deviation of 17.1 at a confidence level of 99.5%. Report ME accurate to one decimal place because the sample statistics are presented with this accuracy. M.E. = Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.
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- The average height of students at a university from an SRS of 14 students gave a standard deviation of 2.5 feet. Construct a 95% confidence interval for the standard deviation of the height of students at the university. Assume normality for the data.A veterinarian wants to estimate the mean weight of a cat in their city. They take a random sample of 19 cats and found a mean of 10.9 lbs with a sample standard deviation of 1 lbs. Find the 95% confidence interval of the mean. Assume that the cat weights are normally distributed. Do not round in between steps. Round answers to at least 4 decimal places.The heights of 9 men were taken with sample mean of 165 cm and standard deviation 5 cm. The heights of 7 Japanese women were taken with sample mean of 150 cm and standard deviation 3 cm. Assuming that the population standard deviations for heights of men and women are both the same, construct the 95% confidence interval for the difference in mean heights between men and women.
- Assume that a sample is used to estimate a population proportion μμ. Find the margin of error M.E. that corresponds to a sample of size 83 with a mean of 72.3 and a standard deviation of 7.3 at a confidence level of 90%.Report ME accurate to one decimal place because the sample statistics are presented with this accuracy.M.E. = Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.In Arkansas, bass fish were collected in 30 different lakes to measure the amount of mercury in the fish. The mean amount of mercury was = 0.6443 with standard deviation s = 0.3606. Estimate the mean amount of mercury in fish in Arkansas lakes using a 95% confidence level. Round your answers to three decimal places, and use four decimal places for any interim calculations. With 95 % confidence, it can said that the true mean amount of mercury in fish in Arkansas lakes is between 0.502 x and 0.786 X Hint: Make sure you round your answer to three decimal places.Assume that a certain data is approximately bell-shaped with mean 91 and standard deviation 6. Use the Empirical Rule to find the percent of the data between 85 and 103. Show all the necessary calculations.
- The scores on a standardazed test are normally distributed with a mean of 115 and standard deviation of 7. What test score is 0.45 standard deviations below the mean?Assume that a sample is used to estimate a population proportion μ. Find the margin of error M.E. that corresponds to a sample of size 50 with a mean of 36.5 and a standard deviation of 11.3 at a confidence level of 80%.Report ME accurate to one decimal place because the sample statistics are presented with this accuracy. M.E. = Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution. Please also state if the assumption of normal distribution was essential for solving this question. You will need to specify the appropriate hypothesis, identify the statistical test, estimate p-value (using R command), and state your conclusion in the context of the problem. For the conclusion, please use the following template. If you reject the null writeSince p-value is [write the value of your p-value] and is less than [write the value of alpha inpercent] we reject the null hypothesis at [write the value of alpha]% significance…
- An advertising agency that serves a major radio station wants to estimate the mean amount of time that the station’s audience spends listening to the radio daily. From past studies, the standard deviation is estimated as 45 minutes. What sample size is needed if the agency wants to be 99% confident of being correct to within ±5 minutes?The ages of a group of 138 randomly selected adult females have a standard deviation of 17.5 years. Assume that the ages of female statistics students have less variation than ages of females in the general population, so leto= 17.5 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want 95% confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population? The required sample size is (Round up to the nearest whole number as needed.)The ponderal index is a measure of overall size similar to a body mass index. The ponderal index of full term newborns follows a normal distribution with a mean of 2.4 kg/m3 and a standard deviation of 0.45 kg/m³. What percentage of full term newborns have a ponderal index greater than 2.5 kg/m³? Use Table 3 to obtain your answer. If necessary, round your final answer to 2 places after the decimal. Do not enter the % symbol, just the number representing the percent.