Find the degrees of freedom and t value given the following sample size and confidence interval. 1. n= 10 a. 95% confidence level b. 98% confidence level c. 99% confidence level
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Q: Find the degrees of freedom and t value given the following sample size and confidence level.
A: Given : Sample size n = 10
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- The 2018 General Social Survey contains information on the number of poor mental health days adults in the US had in the last month. The data are below. Construct the 90% confidence interval for the mean number of poor mental health days for males. Table 1. Mean number of poor mental health days in the last 30 days by sex Male Female Total 3.55 6.78 1408 Mean 3.02 4.06 S.D. 6.43 7.07 678 730 Source: GSS 2018 Your agswer The 2018 General Social Survey contoins information on the nomberetYou are interested in finding a 90% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 10 randomly selected college students. Round answers to 3 decimal places where possible. 1 3 10 10 12 11 10 6 10 3 a. To compute the confidence interval use a Correct distribution. b. With 90% confidence the population mean number of days of class that college students miss is between and days. c. If many groups of 10 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of missed class days and about percent will not contain the true population mean number of missed class days.A sample of freshman takes a reading comprehension test and their scores are summarized below. Sample scores: 16, 8, 8, 6, 9, 11, 13, 9, 10 a. if the mean for the general population on this test is m=12, can you conclude that this sample is significantly different from the population. Test with a=.05 b. Calculate a 90% confidence interval around your mean.
- You are interested in finding a 98% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 14 randomly selected non-residential college students. 8 6 20 26 11 21 23 10 9 23 16 24 23 17 a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean commute for non-residential college students is between and miles. c. If many groups of 14 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of commute miles and about percent will not contain the true population mean number of commute miles.Help with answering true or falseLo A life satisfaction measure is normally distributed with u = 66. Nine individuals with only a primary education are randomly selected to take the life satisfaction measure. Construct and interpret a 90% confidence interval based on the samples results in the table below. 56 57 60 63 68 63 47 45 63
- Assuming that the population is normally distributed, construct a 95% confidence interval for the population mean for each of the samples below. Explain why these two samples produce different confidence intervals even though they have the same mean and range. Sample A: 1 2 4 4 5 5 7 8 Full data set Sample B: 1 2 3 4 5 6 7 8 Question content area bottom Part 1 Construct a 95% confidence interval for the population mean for sample A. enter your response here≤μ≤enter your response hereUse the given degree of confidence and sample data to construct a confidence interval for the population mean u. Assume that the population has a normal distribution. The principal randomly selected six students to take an aptitude test. Their scores were: 88.0 84.1 74.9 83.2 83.7 85.5 Determine a 90% confidence interval for the mean score for all students. Seleccione una: A. 79.49 < µ < 86.98 B. 86.98 < H < 79.49 C. 79.59 < u < 86.88 D. 86.88 < u <79.59Use the given set of points to construct a 95% confidence interval for β1. No need to interpret the results. x 9 6 5 9 6 12 17 11 y 25 18 15 21 21 28 30 24
- You are interested in finding a 90% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 12 randomly selected physical therapy patients. Round answers to 3 decimal places where possible. 7 26 5 10 23 14 14 8 19 6 8 12 a. To compute the confidence interval use a ? z or t distribution. b. With 90% confidence the population mean number of visits per physical therapy patient is between __and ___visits. c. If many groups of 12 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About __percent of these confidence intervals will contain the true population mean number of visits per patient and about __percent will not contain the true population mean number of visits per patient.Researchers studied the mean egg length (in millimeters) for a bird population. After taking a random sample of eggs, they obtained a 95% confidence interval of (45,60). What is the value of the margin of error? Choose the correct answer below. A. 15 mm B. 52.5 mm O c. 7.5 mm O D. 1.9610 The interval estimate is, a 12.2 13.4 b 11.8 13.8 c 11.1 14.5 d 10.0 15.6
