Question

Q.24.A sample of 55 research cotton samples resulted in a sample average percentage elongation of 8.11 and a sample standard deviation of 1.46. Calculate a 95% large-sample CI for the true average percentage elongation μ. (Round your answers to three decimal places.)

What assumptions are you making about the distribution of percentage elongation?
We make no assumptions about the distribution of percentage elongation.
We assume the distribution of percentage elongation is normal with the value of σ unknown.
We assume the distribution of percentage elongation is uniform.
We assume the distribution of percentage elongation is normal with the value of σ known.

Solution :

Given that,

Point estimate = sample mean = = 8.11

sample standard deviation = s = 1.46

sample size = n = 55

Degrees of freedom = df = n - 1 = 55 - 1 = 54

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,54 = 2.005

Margin of error = E = t/2,df * (s /n)

= 2.005 * ( 1.46 / 55)

Margin of error = E = 0.395

The 95% confidence interval estimate of the population mean is,

± E

= 8.11  ± 0.395

= ( 7.715, 8.505 )

We assume the distribution of percentage elongation is normal with the value of σ unknown.

#### Earn Coins

Coins can be redeemed for fabulous gifts.

Similar Homework Help Questions
• ### A sample of 55 research cotton samples resulted in a sample average percentage elongation of 8.16...

A sample of 55 research cotton samples resulted in a sample average percentage elongation of 8.16 and a sample standard deviation of 1.48. Calculate a 95% large-sample CI for the true average percentage elongation p. (Round your answers to three decimal places.) (7777 x) 8.543 What assumptions are you making about the distribution of percentage elongation? We assume the distribution of percentage elongation is normal with the value of e unknown. We assume the distribution of percentage elongation is uniform...

• ### A sample of 54 research cotton samples resulted in a sample average percentage elongation of 8.13...

A sample of 54 research cotton samples resulted in a sample average percentage elongation of 8.13 and a sample standard deviation of 1.47. Calculate a 95% large-sample CI for the true average percentage elongation u. (Round your answers to three decimal places.) What assumptions are you making about the distribution of percentage elongation? We assume the distribution of percentage elongation is normal with the value of o known. We make no assumptions about the distribution of percentage elongation. We assume...

• ### A sample of 57 research cotton samples resulted in a sample average percentage elongation of 8.16 and a sample standard...

A sample of 57 research cotton samples resulted in a sample average percentage elongation of 8.16 and a sample standard deviation of 1.45. Calculate a 95% large-sample CI for the true average percentage elongation u. (Round your answers to three decimal places What assumptions are you making about the distribution of percentage elongation? We assume the distribution of percentage elongation is normal with the value of o known. We make no assumptions about the distribution of percentage elongation. We assume...

• ### A sample of 57 research cotton samples resulted in a sample average percentage elongation of 8.16...

A sample of 57 research cotton samples resulted in a sample average percentage elongation of 8.16 and a sample standard deviation of 1.45. Calculate a 95% large-sample CI for the true average percentage elongation u. (Round your answers to three decimal places.) (7.775 7.775 x 8.545 What assumptions are you making about the distribution of percentage elongation? We assume the distribution of percentage elongation is normal with the value of o known. We make no assumptions about the distribution of...

• ### An article reported that for a sample of 54 kitchens with gas cooking appliances monitored during...

An article reported that for a sample of 54 kitchens with gas cooking appliances monitored during a one week period, the sample mean CO, level (ppm) was 654.16, and the sample standard deviation was 163.53. (a) Calculate and interpret a 95% (two-sided) confidence interval for true average Co, level in the population of all homes from which the sample was selected (Round your answers to two decimal places.) (609.53 x . 098.79 x ) ppm A sample of 55 research...

• ### The Nero Match Company sells matchboxes that are supposed to have an average of 40 matches...

The Nero Match Company sells matchboxes that are supposed to have an average of 40 matches per box, with σ = 8. A random sample of 94 matchboxes shows the average number of matches per box to be 42.2. Using a 1% level of significance, can you say that the average number of matches per box is more than 40? What are we testing in this problem? single mean single proportion (a) What is the level of significance? State the...

• ### The average annual miles driven per vehicle in the United States is 11.1 thousand miles, with...

The average annual miles driven per vehicle in the United States is 11.1 thousand miles, with σ ≈ 600 miles. Suppose that a random sample of 41 vehicles owned by residents of Chicago showed that the average mileage driven last year was 10.9 thousand miles. Does this indicate that the average miles driven per vehicle in Chicago is different from (higher or lower than) the national average? Use a 0.05 level of significance. What are we testing in this problem?...

• ### for these questions I need help on just the questions on all of them where it...

for these questions I need help on just the questions on all of them where it says "what is the value of the sample test statistic?" 24.) The average annual miles driven per vehicle in the United States is 11.1 thousand miles, with σ ≈ 600 miles. Suppose that a random sample of 36 vehicles owned by residents of Chicago showed that the average mileage driven last year was 10.9thousand miles. Does this indicate that the average miles driven per...

• ### The Nero Match Company sells matchboxes that are supposed to have an average of 40 matches per box, with σ = 8. A random...

The Nero Match Company sells matchboxes that are supposed to have an average of 40 matches per box, with σ = 8. A random sample of 94 matchboxes shows the average number of matches per box to be 42.2. Using a 1% level of significance, can you say that the average number of matches per box is more than 40? What are we testing in this problem? single meansingle proportion     (a) What is the level of significance? State the null...

• ### The average annual miles driven per vehicle in the United States is 11.1 thousand miles, with σ ≈ 600 miles. Suppose tha...

The average annual miles driven per vehicle in the United States is 11.1 thousand miles, with σ ≈ 600 miles. Suppose that a random sample of 31 vehicles owned by residents of Chicago showed that the average mileage driven last year was 10.8 thousand miles. Does this indicate that the average miles driven per vehicle in Chicago is different from (higher or lower than) the national average? Use a 0.05 level of significance. What are we testing in this problem?...