Respuesta :
Answer:
a) [tex]58-1.984\frac{8}{\sqrt{100}}=56.41[/tex]
[tex]58+1.984\frac{8}{\sqrt{100}}=59.59[/tex]
So on this case the 95% confidence interval would be given by (56.41;59.59)
b) [tex]62-1.653\frac{11}{\sqrt{200}}=60.71[/tex]
[tex]62+1.653\frac{11}{\sqrt{200}}=63.29[/tex]
So on this case the 90% confidence interval would be given by (60.71;63.29)
c)
[tex]df=n_{1}+n_{2}-2=100+200-2=298[/tex]
[tex]p_v =2*P(t_{(298)}<-3.21)=0.0014[/tex]
So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis
Step-by-step explanation:
Previous concepts
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[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Part a
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\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=100-1=99[/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(0.025,99)".And we see that [tex]t_{\alpha/2}=1.984[/tex]
Now we have everything in order to replace into formula (1):
[tex]58-1.984\frac{8}{\sqrt{100}}=56.41[/tex]
[tex]58+1.984\frac{8}{\sqrt{100}}=59.59[/tex]
So on this case the 95% confidence interval would be given by (56.41;59.59)
Part b
The confidence interval for the mean is given by the following formula:
(1)
In order to calculate the critical value we need to find first the degrees of freedom, given by:
[tex]df=n-1=200-1=199[/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(0.05,199)".And we see that [tex]t_{\alpha/2}=1.653[/tex]
Now we have everything in order to replace into formula (1):
[tex]62-1.653\frac{11}{\sqrt{200}}=60.71[/tex]
[tex]62+1.653\frac{11}{\sqrt{200}}=63.29[/tex]
So on this case the 90% confidence interval would be given by (60.71;63.29)
Part c
For this case the two standard errors are given:
[tex] SE= 1.984\frac{8}{\sqrt{100}}=1.59[/tex]
[tex] SE= 1.653\frac{11}{\sqrt{200}}=1.29[/tex]
[tex]\bar X_1=58[/tex] represent the mean for New Jersey
[tex]\bar X_2=62[/tex] represent the mean for Iowa
[tex]s_1=8[/tex] represent the sample standard deviation for New Jersey
[tex]s_2=11[/tex] represent the sample standard deviation for Iowa
[tex]n_1=100[/tex] sample size for New Jersey
[tex]n_2=200[/tex] sample size for Iowa
t would represent the statistic (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the mean are equal , the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1} = \mu_{2}[/tex]
Alternative hypothesis:[tex]\mu_{1} \neq \mu_{2}[/tex]
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
We can replace in the t formula the result obtained is:
[tex]t=\frac{58-62}{\sqrt{\frac{(8)^2}{100}+\frac{(11)^2}{200}}}}=-3.21[/tex]
Statistical decision
For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{1}+n_{2}-2=100+200-2=298[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{(298)}<-3.21)=0.0014[/tex]
So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
