Respuesta :
Answer:
a) [tex]t=\frac{(134.163 -139.649)-(0)}{\sqrt{\frac{92.857^2}{7}+\frac{98.435^2}{7}}}=-0.107[/tex]
Now we can calculate the degrees of freedom given by:
[tex]df=7+7-2=12[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =2*P(t_{14}<-0.107) =0.917[/tex]
b) The 95% confidence interval would be given by [tex]-116.986 \leq \mu_1 -\mu_2 \leq 106.014[/tex]
Step-by-step explanation:
Data given
Freq. Finite Element Cycle/s Equivalent Plate, Cycle/s
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1 14.58 14.76
2 48.52 49.10
3 97.22 99.99
4 113.99 117.53
5 174.73 181.22
6 212.72 220.14
7 277.38 294.80
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The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 = \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2\neq 0[/tex]
Our notation on this case :
[tex]n_1 =7[/tex] represent the sample size for group 1 ( Finite Element_
[tex]n_2 =7[/tex] represent the sample size for group 2 (Equivalent Plate)
[tex]\bar X_1 =134.163[/tex] represent the sample mean for the group 1 ( Finite Element)
[tex]\bar X_2 =139.649[/tex] represent the sample mean for the group 2 (Equivalent Plate)
[tex]s_1=92.857[/tex] represent the sample standard deviation for group 1 ( Finite Element)
[tex]s_2=98.435[/tex] represent the sample standard deviation for group 2 (Equivalent Plate)
a. Do the data suggest that the two methods provide the same meanvalue for natural vibrationfrequency? Use α = 0.05. Find the P-value.
And the statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}[/tex]
And now we can calculate the statistic:
[tex]t=\frac{(134.163 -139.649)-(0)}{\sqrt{\frac{92.857^2}{7}+\frac{98.435^2}{7}}}=-0.107[/tex]
Now we can calculate the degrees of freedom given by:
[tex]df=7+7-2=12[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =2*P(t_{14}<-0.107) =0.917[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance that difference of means is not different from 0.
b. Find a 95% confidence interval on the mean difference between the two methods
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_1 -\bar X_2 =134.163-139.649=-5.486[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n_1 +n_2 -1=7+7-2=12[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,12)".And we see that [tex]t_{\alpha/2}=2.18[/tex]
Now we have everything in order to replace into formula (1):
[tex]-5.486-2.18\sqrt{\frac{92.857^2}{7}+\frac{98.435^2}{7}}=-116.986[/tex]
[tex]-5.486+2.18\sqrt{\frac{92.857^2}{7}+\frac{98.435^2}{7}}=106.014[/tex]
So on this case the 95% confidence interval would be given by [tex]-116.986 \leq \mu_1 -\mu_2 \leq 106.014[/tex]
