Respuesta :
Answer:
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 ]
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 :
- Sample = [ 418, 421, 421, 422, 425, 428, 431, 435, 437, 438, 445, 447, 448, 453, 458, 462, 465 ]
- 95% confidence interval.
The total number of values given is, n = 17
Mean, [tex]\mu_0=438.47[/tex]
Standard Deviation, [tex]\sigma = 14.79[/tex]
The normal distribution is given by: N (438.47 ; 14.79)
If Cl is 95% then [tex]\alpha[/tex] is 5% and [tex]\alpha /2[/tex] is 2.5%
[tex]\alpha /2 = 0.025[/tex]
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.
[tex]\rm Cl = [\mu_0 \pm t_{\alpha /2};(n-1)\times \dfrac{\sigma}{\sqrt{n} }][/tex]
[tex]\rm Cl = [\mu_0 \pm 2.120\times \dfrac{14.79}{\sqrt{17} }][/tex]
[tex]\rm Cl = [\mu_0 \pm 7.60][/tex]
Cl = [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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https://brainly.com/question/2561151
