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 ]

total amount to be paid = $19.5

step-by-step explanation:

given ,

original price = $25.99

discount = 25%

discount amount = 25% of original price

discount amount = 25/100   * 25.99

discount amount = $6.49

hence , the total amount to be paid = original price - discount amount

                                                            = $25.99 - $6.49

                                                            =$19.5

total amount to be paid = $19.5