calculate a two-sided 95% confidence interval for true average degree of polymerization. (round your answers to two decimal places.)

Respuesta :

95% confidence interval for true average degree of polymerization is  [430.87 ; 446.07] and this interval suggest that 441 is a plausible value for true average degree of polymerization and also this interval does not suggest that 451 is a plausible value.

Given :

  1. Sample = [ 418, 421, 421, 422, 425, 428, 431, 435, 437,  438, 445, 447, 448, 453, 458, 462, 465 ]
  2. 95% confidence interval.

The total number of values given is, n = 17

Mean, μ₀ = 438.47

Standard Deviation, σ = 14.79

The normal distribution is given by: N (438.47 ; 14.79)

If Cl  is 95% then  is 5% and  is 2.5%

α/2 = 0.025

Now, use t-statistics distribution with (n -1) degree of freedom

                         df = 16

So, the t score for 0.025 and 16 s from t-table 2.120.

CI = [ μ₀ ± [tex]t_{\alpha /2}[/tex] ; (n-1) × σ/ √n]

⇒ CI = [μ₀ ± 2.120 × 14.79/√17

⇒ CI = [ μ₀ ± 7.60]

⇒ CI = 430.87 ; 446.07]

Yes, the interval suggests that 441 is a plausible value for true average degree of polymerization.

No, the interval does not suggest that 451 is a plausible value.

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