Respuesta :
Answer:
a. 1. Two measurements (A and B) are taken on the same round.
b. No, there is not enough evidence to support the claim that the difference d is significantly different from 0.
c. The 99% confidence interval for the mean difference is (-8.76, 2.26).
This means that we have 99% confidence that the true mean difference of measurements between device A minus device B is within -8.76 and 2.26. As the value 0 is within the interval, we can not claim that there is a significant difference between the population difference between the measurements of both devices: they could be calibrated (mean difference equal to 0) or not.
Step-by-step explanation:
a) This is a matched-pairs because the samples are related: the n-th round is measured with each device, and one measurement is in the sample A and the other is in sample B.
b) For matched pairs data, a new variable "d" is calculated as the difference between each pair, and tested if d is significantly different from 0.
The sample data for d is [9.9, 9.9, -1.6, -1.6, 8.4, 8.4, -1.1, -1.1, 8.7, 8.7, -4.8, -4.8 ].
The mean and standard deviation for d are:
[tex]M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{12}((-9.9)+(-9.9)+1.6+. . .+4.8)\\\\\\M=\dfrac{-39}{12}\\\\\\M=-3.25\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{11}((-9.9-(-3.25))^2+(-9.9-(-3.25))^2+(1.6-(-3.25))^2+. . . +(4.8-(-3.25))^2)}\\\\\\s=\sqrt{\dfrac{415.39}{11}}\\\\\\s=\sqrt{37.76}=6.15\\\\\\[/tex]
Then, we perform an hypothesis matched pair t-test.
This is a hypothesis test for the population mean.
The claim is that the difference d is significantly different from 0.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu_d=0\\\\H_a:\mu_d\neq 0[/tex]
The significance level is 0.01.
The sample has a size n=12.
The sample mean is M=-3.25.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.15.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.15}{\sqrt{12}}=1.775[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{-3.25-0}{1.775}=\dfrac{-3.25}{1.775}=-1.831[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=12-1=11[/tex]
This test is a two-tailed test, with 11 degrees of freedom and t=1.831, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=2\cdot P(t<-1.831)=0.094[/tex]
As the P-value (0.094) is bigger than the significance level (0.01), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the difference d is significantly different from 0.
c. We have to calculate a 99% confidence interval for the mean.
The t-value for a 99% confidence interval and 11 degrees of freedom is t=3.106.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=3.106 \cdot 1.775=5.51[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = -3.25-5.51=-8.76\\\\UL=M+t \cdot s_M = -3.25+5.51=2.26[/tex]
The 99% confidence interval for the mean difference is (-8.76, 2.26).
This means that we have 99% confidence that the true mean difference of measurements between device A minus device B is within -8.76 and 2.26. As the value 0 is within the interval, we can not claim that there is a significant difference between the population difference between the measurements of both devices: they could be calibrated (mean difference equal to 0) or not.
