Question

In: Statistics and Probability

Find the requested confidence interval. A random sample of 30 long distance runners aged 20-25 was...

Find the requested confidence interval. A random sample of 30 long distance runners aged 20-25 was selected from a running club. The mean of resting heart rates (in beats per minute) of the runners taken in this sample is 68.07 Estimate the mean resting heart rate for the population of long distance runners aged 20-25. Assuming ? = 8.03, give the 95.44% confidence interval for the population mean.

Solutions

Expert Solution

Solution :

Given that,

= 68.07

= 8.03

n = 30

At 95.44% confidence level the z is ,

  = 1 - 95.44% = 1 - 0.9544= 0.0456

/ 2 = 0.0456 / 2 = 0.0228

Z/2 = Z0.0228 = 1.999

Margin of error = E = Z/2* (/n)

= 1.999 * (8.03 / 30 ) = 2.93

At 95.44% confidence interval estimate of the population mean is,

- E < < + E

68.07 - 2.93 < < 68.07 + 2.93

65.14 < < 71.00

(65.14 , 71.00)


Related Solutions

Given a random sample of size 322. Find a 99% confidence interval for the population proportion...
Given a random sample of size 322. Find a 99% confidence interval for the population proportion if the number of successes was 168. (Use 3 decimal places.) lower limit upper limit
Use the given degree of confidence and sample data to find a confidence interval for the...
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Weights of men: 90% confidence; n = 14, = 155.7 lb, s = 13.6 lb A. 11.0 lb < σ < 2.7 lb B. 10.1 lb < σ < 19.1 lb C. 10.4 lb < σ < 20.2 lb D. 10.7 lb < σ < 17.6 lb
Find the requested confidence interval. The data below consists of the test scores of 32 students....
Find the requested confidence interval. The data below consists of the test scores of 32 students. Construct a 95.44% confidence interval for the population mean. 88 70 67 95 99 72 86 60 57 92 90 88 70 73 88 97 109 63 76 71 97 88 70 79 54 80 83 92 91 66 90 90
in a random sample of 500 people aged 20-24, 110 were smokers. in a random sample...
in a random sample of 500 people aged 20-24, 110 were smokers. in a random sample of 450 people aged 25-29, 65 were smokers. test the claim that the proprtion of smokers age 20-24 is higher than the proportion of smokers age 25-29. use a significance level of 0.01. please show work!
Use technology and the given confidence level and sample data to find the confidence interval for...
Use technology and the given confidence level and sample data to find the confidence interval for the population mean mu. Assume that the population does not exhibit a normal distribution. Weight lost on a diet: 95 % confidence n equals 51 x overbar equals 4.0 kg s equals 6.9 kg What is the confidence interval for the population mean mu?? nothing kgless thanmuless than nothing kg ?(Round to one decimal place as? needed.)
Use the confidence level and sample data to find a confidence interval for estimating the population...
Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your answer to the same number of decimal places as the sample mean. Test scores: n = 81,  = 69.0, σ = 4.6; 98% confidence
Use the confidence level and sample data to find a confidence interval for estimating the population...
Use the confidence level and sample data to find a confidence interval for estimating the population mean mu. A sociologist develops a test to measure attitudes towards public transportation, and 35 randomly selected subjects are given the test . If the sample mean was 81.2 with the sample standard deviation was 24.1, construct a confidence interval to estimate the population mean if confidence level is 95% Initial Data : E(margin of error result value) = Cl(population confidence inerval result value...
Use technology and the given confidence level and sample data to find the confidence interval for...
Use technology and the given confidence level and sample data to find the confidence interval for the population mean u. Weight lost on a diet: 98% Confidence n=51 x=2.0 kg    s=4.2 kg What is the confidence interval for the population mean u? ___ kg < u < ___ kg
Use the confidence level and sample data to find a confidence interval for estimating the population...
Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your answer to the same number of decimal places as the sample mean. Test scores: n = 104,  = 95.3, σ = 6.5; 99% confidence Group of answer choices a. 94.2 < μ < 96.4 b. 93.7 < μ < 96.9 c. 93.8 < μ < 96.8 d. 94.1 < μ < 96.5
Use technology and the given confidence level and sample data to find the confidence interval for...
Use technology and the given confidence level and sample data to find the confidence interval for the population mean. Assume that the population does not exhibit a normal distribution. Weight lost on a diet: 99% confindence n=41 dash above x = 4.0 kg s=5.9 kg What is the confidence interval for the population mean μ ? __kg < mean < __kg Is the confidence interval affected by the fact that the data appear to be from a population that is...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT