Respuesta :
Answer:
The 98% confidence interval would be given by [tex]-1.313 \leq \mu_1 -\mu_2 \leq 0.933[/tex]
Step-by-step explanation:
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".
[tex]\bar X_1 =4.83[/tex] represent the sample mean 1
[tex]\bar X_2 =5.02[/tex] represent the sample mean 2
n1=32 represent the sample 1 size
n2=35 represent the sample 2 size
[tex]s_1 =1.65[/tex] population standard deviation for sample 1
[tex]s_2 =2.18[/tex] population standard deviation for sample 2
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
Solution to the problem
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 =4.83-5.02=-0.19[/tex]
Now we need to find the degrees of freedom given by:
[tex]df=n_1 +n_2 -2= 32+35-2=65[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,65)".And we see that [tex]t_{\alpha/2}=2.39[/tex]
Now we have everything in order to replace into formula (1):
[tex]-0.19-2.39\sqrt{\frac{1.65^2}{32}+\frac{2.18^2}{35}}=-1.313[/tex]
[tex]-0.19+2.39\sqrt{\frac{1.65^2}{32}+\frac{2.18^2}{35}}=0.933[/tex]
So on this case the 98% confidence interval would be given by [tex]-1.313 \leq \mu_1 -\mu_2 \leq 0.933[/tex]
