Respuesta :
Answer: (a) No, the data suggests it is possible to reject the hypothesis which states that the 2 methods provide same mean value.
(b) (-10.957,-0.063)
Step-by-step explanation:
(a) Hypotheses for this data are:
[tex]H_{0}[/tex]: [tex]\mu_{1} = \mu_{2}[/tex]
[tex]H_{a}: \mu_{1} \neq \mu_{2}[/tex]
First find the differences in values:
1 => -0.31
2 => -0.63
3 => -2.85
4 => -3.46
5 => -6.43
6 => -7.52
7 => -17.39
Now, find the mean and standard deviation of the differences:
mean = [tex]\frac{-0.31+(-0.63)+...+(-17.39)}{7}[/tex] = - 5.51
sd = [tex]\sqrt{\frac{(-0.31-(-5.51))^{2}+...+(-17.39-(-5.51))^{2}}{7-1} }[/tex] = 5.89
The value of test statistics is:
t = [tex]\frac{mean}{\frac{s}{\sqrt{n} } }[/tex] = [tex]\frac{-5.51}{\frac{5.89}{\sqrt{7} } }[/tex] = - 2.4750
Analysing Student's T distribution, at a df = 7-1 = 6:
p-value = 0.025*2 = 0.05
To reject the null hypothesis, p-value must be less or equal than α. Since they are equal, reject the null hypothesis, i.e., reject the claim suggesting the 2 methods provide the same mean value for natural vibration frequency.
(b) For a CI = 95%:
t-score for α = 0.025 and df = 6 is 2.447.
mean ± [tex]t.\frac{s}{\sqrt{n} }[/tex]
-5.51 ± 2.447.[tex]\frac{5.89}{\sqrt{7} }[/tex]
-5.51 ± 5.4471
lower limit: -5.51 - 5.4471 = - 10.957
upper limit: -5.51 + 5.4471 = - 0.063
The interval on the mean difference is (-10.957,-0.063)
