Respuesta :
Answer:
a) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]
[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]
b) The 90% confidence interval would be given by (63.330;81.070)
[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]
Step-by-step explanation:
1) Previous concepts and notation
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Let put some notation
x=left arm , y = right arm
x: 175 169 182 146 144
y: 102 101 94 79 79
The first step is define the difference [tex]d_i=x_i-y_i[/tex], that is given so we have:
d: 73, 68, 88, 67, 65
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]
2) Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}[/tex] (1)
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=5-1=4[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that [tex]t_{\alpha/2}=2.13[/tex]
Now we have everything in order to replace into formula (1):
[tex]72.2-2.13\frac{9.311}{\sqrt{5}}=63.330[/tex]
[tex]72.2+2.13\frac{9.311}{\sqrt{5}}=81.070[/tex]
So on this case the 90% confidence interval would be given by (63.330;81.070)
[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]
