A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.63 and a sample standard deviation of 0.79. calculate a 95% large-sample ci for the true average percentage elongation μ. (give answers accurate to 2 decimal places.)

Question

06-08-2017
Mathematics

contestada

A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.63 and a sample standard deviation of 0.79. calculate a 95% large-sample ci for the true average percentage elongation μ. (give answers accurate to 2 decimal places.)

Respuesta :

Otras preguntas

Nancy is knitting a scarf. It takes her 0. 2 of an hour to knit each inch of the scarf. She wants the length of the scarf to be 60 inches. How many hours will i

Aqueous lead (II) nitrate, Pb(NO3)2 undergoes a double displacement reaction with aqueous sodium chloride, NaCl, in which a precipitate forms. If the precipit

A sudden storm threatens the property of Marina Bay LLC. Ngoc, Marina Bay’s sales agent, is on site, but lacks the express or implied authority to bind the firm

true or false: requestor identity and timestamps of server requests are types of supplemental analytics information.

A bowling ball that weighs 150kg is tossed down a bowling love and hits the Pins with 1500N of Force what Was the acceleration of the bowling ball moving down t

Which philosophy thought people should live simply and naturally?

The Kw for water at 0 ∘C is 0.12×10^−14. Calculate the pH of a neutral aqueous solution at 0 ∘C.pH=Is a pH=7.25 solution acidic, basic, or neutral at 0 ∘C?a. ac

. how much of a 750-gram sample of potassium -42 is left after 62 hours? (round to two decimal places)

place the following events in order starting with the vibration of the oval window and ending with the depolarization of hair cells1) increased pressure in peri

write an oz program considering let-expressions (extending the previous expression evaluators from past homeworks).

barnuts barnuts · Answer 1 · 2017-08-17T09:21:39+02:00

To solve for the confidence interval for the true average percentage elongation, we use the z statistic. The formula for confidence interval is given as:

Confidence interval = x ± z σ / sqrt (n)

where,

x = the sample mean = 8.63

σ = sample standard deviation = 0.79

n = number of samples = 56

From the standard distribution tables, the value of z at 95% confidence interval is:

z = 1.96

Therefore substituting the known values into the equation:

Confidence interval = 8.63 ± (1.96) (0.79) / sqrt (56)

Confidence interval = 8.63 ± 0.207

Confidence interval = 8.42, 8.84