Round to three decimal places as needed.) Part 4 The confidence interval estimate of sigma is response here mgless thansigmaless than enter your enter your response here mg. (Round to two decimal places as needed.)
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Q: Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t Suppose a small…
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- For the table below. Determine the regression coefficient (with confidence interval), and the adjustment line:Use the sample information = 37, 0=5, n=15 to calculate the following confidence intervals for u assuming the sample is from a normal population. (a) 90 percent confidence. (Round your answers to 4 decimal places.) The 90% confidence interval is from to (b) 95 percent confidence. (Round your answers to 4 decimal places.) The 95% confidence interval is from to (c) 99 percent confidence. (Round your answers to 4 decimal places.) es The 99% confidence interval is from to (d) Describe how the intervals change as you increase the confidence level. O The interval gets narrower as the confidence level increases. O The interval gets wider as the confidence level decreases. O The interval gets wider as the confidence level increases. O The interval stays the same as the confidence level increases. 4 of 10 NEht > < Prev ere to searchUse SALT to find the multiplier t* for calculating a confidence interval for a single population mean in each of the following situations. (Round your answers to two decimal places.) (a) n = 27; confidence level = 0.98 (b) n = 6; confidence level = 0.90 (c) n = 6; confidence level = 0.99
- # 34- In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before−after) in their levels of LDL cholesterol (in mg/dL) have a mean of 4.9 and a standard deviation of 17.5. Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? What is the confidence interval estimate of the population mean μ? _____mg/dL<μ<_____mg/dL (Round to two decimal places as needed.) What does the confidence interval suggest about the effectiveness of the treatment? A. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. B. The confidence interval limits contain 0, suggesting that the garlic treatment…Use the sample information I= 37, 0 = 5, n= 15 to calculate the following confidence intervals for u assuming the sample is from a normal population. (a) 90 percent confidence. (Round your answers to 4 decimal places.) The 90% confidence interval is from to (b) 95 percent confidence. (Round your answers to 4 decimal places.) The 95% confidence interval is from to (c) 99 percent confidence. (Round your answers to 4 decimal places.) The 99% confidence interval is from to (d) Describe how the intervals change as you increase the confidence level. O The interval gets narrower as the confidence level increases. O The interval gets wider as the confidence level decreases. O The interval gets wider as the confidence level increases. O The interval stays the same as the confidence level increases. |出 NGMt > < Prev 4 of 10 6:5 2/28 ere to searchX Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) below. f a. What is the number of degrees of freedom that should be used for finding the critical value /? df = 0 (Type a whole number.) b. Find the critical value a/2 corresponding to a 95% confidence level. a/2= (Round to two decimal places as needed.) c. Give a brief general description of the number of degrees of freedom. D 5373255.docx OA. The number of degrees of freedom for a collection of sample data is the number of sample values that are determined after certain restrictions have been imposed on all data values. OB. The number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values. OC. The number of degrees of freedom for a collection of sample data is the total number of sample values. OD. The number of degrees of freedom for a collection…
- We have the survey data on the body mass index (BMI) of 659 young women. The mean BMI in the sample was x = 27.7. We treated these data as an SRS from a Normally distributed population with standard deviation σ = 7.5. Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence. (Round your answers to two decimal places.)Use the sample Information a = 36, o = 6, n= 11 to calculate the following confidence Intervals for u assuming the sample is from a normal population. (a) 90 percent confidence. (Round your answers to 4 decimal places.) The 90% confidence Interval Is from to (b) 95 percent confidence. (Round your answers to 4 decimal places.) The 95% confidence interval is from to (c) 99 percent confidence. (Round your answers to 4 decimal places.) The 99% confidence Interval is from ton1=16 x‾1=124 s1=35 n2=12 x‾2=86 s2=31 Use this data to find the 98% confidence interval for the true difference between the population means. Assume that the population variances are not equal and that the two populations are normally distributed. Copy Data Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places. Step 3 of 3 : Please also find the high and low endpoints and round to four decimal points.
- see belowAllen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t Suppose a small group of 20 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with o = 0.32 gram. (a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) O normal distribution of weights Oo is known O n is large O o is unknown O uniform distribution of weights (c) Interpret your results in the context of this problem. O The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. O we are 80% confident that the true average weight of Allen's hummingbirds falls…Use the sample information I= 36, 0 = 4, n= 13 to calculate the following confidence intervals for u assuming the sample is from a normal population. (a) 90 percent confidence. (Round your answers to 4 decimal places.) The 90% confidence interval is from to (b) 95 percent confidence. (Round your answers to 4 decimal places.) The 95% confidence interval is from to (c) 99 percent confidence. (Round your answers to 4 decimal places.) The 99% confidence interval is from to (d) Describe how the intervals change as you increase the confidence level. O The interval gets narrower as the confidence level increases. O The interval gets wider as the confidence level decreases. The interval gets wider as the confidence level increases. O The interval stays the same as the confidence level increases. 3:32 PM 10/25/2020 %23 O e to search