Respuesta :
Answer:
a. Best estimate for the mean family size for the population of all families in a country is 16.3571428571
b. 95% confidence interval for the estimate is 16.3571428571±0.7106240840. That is between 15.6465187731 and 17.0677669411
c. The sample is unlikely to be very representative of all families in a country.
Step-by-step explanation:
a. mean family size for the population of all families in a country can be estimated as sample mean, and calculated as:
[tex]\frac{15+ 16+ 19+ 18+ 17+ 15+ 16+ 16+ 15+ 16+ 15+ 16+ 17+ 18 }{14}[/tex] ≈ 16.3571428571
b. 95% confidence interval for the estimate can be calculated using the equation CI=M± [tex]\frac{t*s}{\sqrt{N} }[/tex] where
- M is the mean estimate (16.3571428571)
- t is the corresponding statistic in the 95% confidence level and with 14 degrees of freedom (2.160)
- s is the standard deviation of the sample (1.2309777100)
- N is the sample size (14)
s can be calculated as square root of mean squared differences from the sample mean.
Thus 95% CI=16.3571428571± [tex]\frac{2.160*1.2309777100}{\sqrt{14} }[/tex] ≈ 16.3571428571±0.7106240840
The sample is unlikely to be very representative of all families in a country because as the sample data increases, estimate will improve.
Answer:
a) [tex]\bar X = \frac{\sum_{i=1}^n X_i }{n}[/tex]
[tex]\bar X=16.357[/tex] represent the sample mean
b) The 95% confidence interval would be given by (15.620;17.094)
c) The concept of very representative not exists. A asmaple can be representative or not of a population. So then based on this the best option is:
The sample is unlikely to be representative of all families in a country.
The reason is because we have just a sample of 14 if we want to be more confidence about the true population mean we can increase the sample size in order to have a better estimation.
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".
The data given : 15, 16, 19, 18, 17, 15, 16, 16, 15, 16, 15, 16. 17, 18
We can calculate the mean and the sample deviation with the following formulas:
[tex]\bar X = \frac{\sum_{i=1}^n X_i }{n}[/tex]
[tex] s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X=16.357[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=1.277[/tex] represent the sample standard deviation
n=14 represent the sample size
Part a
The best estimate for the mean is the average given by:
[tex]\bar X = \frac{\sum_{i=1}^n X_i }{n}[/tex]
[tex]\bar X=16.357[/tex] represent the sample mean
Part b
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)
The degrees of freedom are given by:
[tex] df = n-1= 14-1 = 13[/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(0.025,13)".And we see that [tex]t_{\alpha/2}=\pm 2.16[/tex]
Now we have everything in order to replace into formula (1):
[tex]16.357-2.16\frac{1.277}{\sqrt{14}}=15.620[/tex]
[tex]16.357+2.16\frac{1.277}{\sqrt{14}}=17.094[/tex]
So on this case the 95% confidence interval would be given by (15.620;17.094)
Part c
The concept of very representative not exists. A asmaple can be representative or not of a population. So then based on this the best option is:
The sample is unlikely to be representative of all families in a country.
The reason is because we have just a sample of 14 if we want to be more confidence about the true population mean we can increase the sample size in order to have a better estimation.
