Homework Answers
here estimated mean=(233.798+230.070)/2=231.934
margin of error =(233.798-230.070)/2=1.864
therefore margin of error for 99% CI =1.864*t0.01/t0.05 =1.864*2.776/4.604 =1.124
hence 99% conifidence interval =231.934 -/+ 1.124 =( 230.810 , 233.058 )
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A journal article reports that a sample of size 5 was used as a basis for...
A journal article reports that a sample of size 5 was used as a basis for...
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.863, 233.605). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? [Hint: Use the center of the interval and its width to determine x and s.] (Round...
A journal article reports that a sample of size 5 was used as a basis for...
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.866, 233.602). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? [Hint: Use the center of the interval and its width to determine x and s.] (Round...
Answer the following question A journal article reports that a sample of size 5 was used...
Answer the following question
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (230.061, 233.807). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? (Hint: Use the center of the interval and its width to determine...
Answer the following questions A journal article reports that a sample of size S was used...
Answer the following questions
A journal article reports that a sample of size S was used as a basis for calculating a 95% C1 for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (230.061, 233.807). You decide that a confidence level of 99 % is more appropriate than the 95 % level used. What are the limits of the 99 % interval? [Hint: Use the center of the interval and its...
Question 4:14 pts) A joumal article sample of sample variance of 2.268 was used as natural...
Question 4:14 pts) A joumal article sample of sample variance of 2.268 was used as natural frequency (Hz) of delamin same sample data. The resulting intervals were (230.2, 233.07) and (229.11, 234.16) reports that a sample of size 5 with a sample mean of 231.634 and a a basis for calculating two confidence intervals for the true average ated beams of a certain type. Both intervals were calculated from the (a) Determine the confidence levels of the two intervals. You...
Please Help with BOTH 1) 2) An article reported that for a sample of 56 kitchens...
Please Help with BOTH
1)
2)
An article reported that for a sample of 56 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 163.36. (a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.) ppm Interpret the resulting interval. We are...
An article reported that for a sample of 54 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was...
An article reported that for a sample of 54 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 162. a) Calculate and interpret a 95% two sided answers to two decimal places.) confidence interval for true average CO2 level n the population of a homes from hich the sample as selected Round your ppm Interpret the resulting interval o we are 95% confident that the...
An article reported that for a sample of 59 kitchens with gas cooking appliances monitored during...
An article reported that for a sample of 59 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 164.04. (a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.) (611.41 696.91 pm Interpret the resulting interval. We are 95% confident that this...
An article investigated the consumption of caffeine among women. A sample of 59 women were asked...
An article investigated the consumption of caffeine among women. A sample of 59 women were asked to monitor their caffeine intake over the course of one day. The mean amount of caffeine consumed in the sample of women was 237.999 mg with a standard deviation of 218.428 mg. In tite article, researchers would like to include a 99% confidence interval. The values below are t critical point corresponding to 90%, 95%, 98%, and 99% confidence, which critical point corresponds to...
A random sample of 43 observations is used to estimate the population variance. The sample mean...
A random sample of 43 observations is used to estimate the population variance. The sample mean and sample standard deviation are calculated as 68.5 and 3.1, respectively. Assume that the population is normally distributed. (You may find it useful to reference the appropriate table: chi-square table or F table) a. Construct the 95% interval estimate for the population variance. (Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) Confidence interval to b. Construct...
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.863, 233.605). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? [Hint: Use the center of the interval and its width to determine x and s.] (Round...
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.866, 233.602). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? [Hint: Use the center of the interval and its width to determine x and s.] (Round...
Answer the following question
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (230.061, 233.807). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? (Hint: Use the center of the interval and its width to determine...
Answer the following questions
A journal article reports that a sample of size S was used as a basis for calculating a 95% C1 for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (230.061, 233.807). You decide that a confidence level of 99 % is more appropriate than the 95 % level used. What are the limits of the 99 % interval? [Hint: Use the center of the interval and its...
Question 4:14 pts) A joumal article sample of sample variance of 2.268 was used as natural frequency (Hz) of delamin same sample data. The resulting intervals were (230.2, 233.07) and (229.11, 234.16) reports that a sample of size 5 with a sample mean of 231.634 and a a basis for calculating two confidence intervals for the true average ated beams of a certain type. Both intervals were calculated from the (a) Determine the confidence levels of the two intervals. You...
Please Help with BOTH
1)
2)
An article reported that for a sample of 56 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 163.36. (a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.) ppm Interpret the resulting interval. We are...
An article reported that for a sample of 54 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 162. a) Calculate and interpret a 95% two sided answers to two decimal places.) confidence interval for true average CO2 level n the population of a homes from hich the sample as selected Round your ppm Interpret the resulting interval o we are 95% confident that the...
An article reported that for a sample of 59 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 164.04. (a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.) (611.41 696.91 pm Interpret the resulting interval. We are 95% confident that this...
An article investigated the consumption of caffeine among women. A sample of 59 women were asked to monitor their caffeine intake over the course of one day. The mean amount of caffeine consumed in the sample of women was 237.999 mg with a standard deviation of 218.428 mg. In tite article, researchers would like to include a 99% confidence interval. The values below are t critical point corresponding to 90%, 95%, 98%, and 99% confidence, which critical point corresponds to...
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