Sample= 38 Sample mean 215 mg Population standard deviation is 24mg Give 99% confidence interval for M, Population mean
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Sample= 38
Sample
Population standard deviation is 24mg
Give 99% confidence interval for M, Population mean
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- estimate mean level of education of unemployed youth less than or equal to 21. standard deviation is 6.7 years 99% accuracy mean is within 2 years. sample sizeData on the numbers of hospital admissions resulting from motor vehicle crashes are given below for Friday’s on the 6th of a month and fridays on the following 13th of the same month. Paired sample data is a simple random sample and the differences have a distribution that is normal. Construct 95% confidence interval to estimate the mean of the population of differences between hospital admissions. 6th: 6,3,7,5,6 13th: 15,14,12,15,12 Find the 95% confidence interval, test statistic and p valueWeights of whole walnuts are normally distributed with mean 9.2 grams and standard deviation 0.104 grams. What proportion of walnuts weigh less than 8.9722 grams? What proportion of walnuts weigh more than 9.0315 grams? Use the closest value in the normal table, and show all four decimals.
- Assum a sample is used to estimate a populatation mean. Find the 95% confidence interval for a sample size of 41 with a mean of 64.2 and a standard deviation of 13.4. Enter answer as an open interval (i.e. parentheses) accurate to 3 decimal places.edu/ultra/courses/_313006_1/cl/outline * Question Completion Status: QUESTION 1 EPA fuel economy estimates for automobile models tested recently predicts a mean of 24 mpg and a standard deviation of 6 mpg. Assume that a Normal model applies, N(24, 6). What proportion of automobiles would you expect to have gas mileage of more than 27 mpg? a. 0.6915 b. 0.0013 Oc C. 0.5000 O d. 0.3085 QUESTION 2You want to estimate the mean weight of quarters in circulation. A sample of 40 quarters has a mean weight of 5.626 grams and a standard deviation of 0.059 gram. Use a single value to estimate the mean weight of all quarters. Also, find the 95% confidence interval for the average weight of all quarters. The estimate for the mean weight of all quarters is (Round to three decimal places as needed.) an example Get more help - grams. Clear all
- nd the 30th and 98th percentile values cf a normal variable with Mean of 12_and Variance of 30 Anot This I NEED PARTBI b), Suppose we sample 3 items from this population. Find the variables that correspond to the central 85% Confident Interval range of the sum of these 3 samples.erproduction of uric acid in the body can be an Indication of cell breakdown. This may be an advance indication of illness such as go adult male patlent has taken eleven blood tests for uric acid. The mean concentration was x = 5.30 mg/dl. The distribution of uric rmal, with o = 1.95 mg/dI. a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of ei lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) normal distribution of uric acid uniform distribution of uric acid o is known On is large Oo is unknown (c) Interpret your results in the context of this problem. O The probability that this interval contains the true average uric acid level for this patient is 0.95. O We are 5% confident that the true uric acid level for this patient falls within this interval. O We are 95% confident that the true uric acid level for this patient falls within…Pls help
- Please do not give solution in image format thanku A dietitian wants to know the average time spent on breakfast in a primary school. The dietitian randomly samples 16 students and finds that the average is 15.8 minutes with a standard deviation of 2.31 minutes. Assume that the distribution of the time spent on the breakfast is normally distributed. The dietitian finds a 90% confidence interval for this sample is (14.788, 16.812). Select one or more: a. The margin of error is 1.012. b.The margin of error is 2.024 c. The margin of error is 0.950. d. We believe that the true mean time spent on breakfast in this primary school is between 14.788 and 16.812 minutes. e. If we take many other samples from this population, 90% of them will have a sample mean that is between 14.788 and 16.812. f. There is a 90% chance that the true mean is between 14.788 and 16.812 minutes.Only test statistic pleasePlease do not give solution in image format thanku If mean=10, standard deviation= 2, then 95% of the scores in the sample fall between: 8 and 12 10 and 12 6 and 14