An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: 418 421 421 422 425 428 431 435 437
438 445 447 448 453 458 462 465
(c) Calculate a two-sided 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.) Note that it is plausible that the given sample observations were selected from a normal distribution and there are no outliers.

(___ , ___)

Does the interval suggest that 441 is a plausible value for true average degree of polymerization?

Yes or No

Does the interval suggest that 451 is a plausible value?

Yes or No

Step-by-step explanation:

Form a set of values we get

n = 17

And with the help of a calculator

μ₀ = 438,47

σ  = 14,79

Normal Distribution is :  N ( 438,47 ; 14,79 )

c)

CI = 95 % means  α = 5 %    α/2 = 2,5 %    α/2 = 0,025

and as n < 30  we should use t-student distribution with n -1 degree of freedom  df = 16.  t score for 0,025 and 16 s from t-table 2,120

By definition:

CI = [  μ₀  ±  t α/2 ; n-1 * σ/√n ]

CI = [  μ₀  ± 2,120* 14,79/√17 ]

CI = [  μ₀  ±  7,60 ]

CI = [ 438,47 ± 7,60 ]

CI = [  430,87 ; 446,07 ]

29.68

work:

okay so to solve this problem, you wanna start by finding 20% of 35. to start, change 20% into a decimal, 0.20, and multiply by 35.

35 x 0.2 = 7

so, if you are getting 20% off, that means you are getting $7 off, so subtract 7 from 35.

35 - 7 = 28.

okay, so that's the sale price, but we still have to add the tax. convert 6% into a decimal (0.06) and multiply by the total.

28 x .06 = 1.68

lastly, add the sales tax onto the jean price to get the total.

28 + 1.68 = 29.68

i hope this helped and have a great day! if you need any further assistance with this specific question or i made any mistakes, please let me know!