Respuesta :
Answer:
a) Point estimate p=0.39
The 95% confidence interval for the population proportion is (0.369, 0.411).
b) Point estimate M=10.9 cigarretes a day.
The 95% confidence interval for the mean is (10.84, 10.96).
c) 27.95 cigarretes per day.
Step-by-step explanation:
The question is incomplete:
The sample data is not attached.
We can work with a random and representative sample where, from 2000 Russians interviewed, 780 are smokers.
Out of this 780 smokers, 550 smoke a package a day, 150 smoke two packages a day and 80 smoke three packages a day. The packages have 20 cigarettes each.
a) We have to calculate a 95% confidence interval for the proportion.
The score is X=780, with a sample size n=2000.
The point estimate for the sample population is the sample proportion and has a value of p=0.39.
[tex]p=X/n=780/2000=0.39[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.39*0.61}{2000}}\\\\\\ \sigma_p=\sqrt{0.000119}=0.011[/tex]
The critical z-value for a 95% confidence interval is z=1.96.
The margin of error (MOE) can be calculated as:
[tex]MOE=z\cdot \sigma_p=1.96 \cdot 0.011=0.021[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=p-z \cdot \sigma_p = 0.39-0.021=0.369\\\\UL=p+z \cdot \sigma_p = 0.39+0.021=0.411[/tex]
The 95% confidence interval for the population proportion is (0.369, 0.411).
b) The point estimate for the mean annual per capita consumption of cigarettes can be calculated as:
[tex]M_X=\dfrac{\sum n_iX_i}{n}=\dfrac{1220\cdot0+550\cdot 20+150\cdot 40+80\cdot 60}{2000}\\\\\\M_X=\dfrac{0+11000+6000+4800}{2000}=\dfrac{21800}{2000}=10.9[/tex]
The standard deviation can be calculated as:
[tex]s_X=\sqrt{\dfrac{\sum n_i(X_i-M_x)^2}{n-1}}\\\\\\s_X=\sqrt{\dfrac{1220(0-10.9)^2+550(20-10.9)^2+150(40-10.9)^2+80(60-10.9)^2}{1999}}\\\\\\s_X=\sqrt{\dfrac{118.81+82.81+846.81+2410.81}{1999}}=\sqrt{\dfrac{3459.24}{1999}}=\sqrt{1.73}\approx1.32[/tex]
We have to calculate a 95% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=10.9.
The sample size is N=2000.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{1.32}{\sqrt{2000}}=\dfrac{1.32}{44.721}=0.03[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=2000-1=1999[/tex]
The t-value for a 95% confidence interval and 1999 degrees of freedom is t=1.961.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=1.961 \cdot 0.03=0.06[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 10.9-0.06=10.84\\\\UL=M+t \cdot s_M = 10.9+0.06=10.96[/tex]
The 95% confidence interval for the mean is (10.84, 10.96).
c. Only for the proportion of smokers, the expected value for the number of cigarretes smoked per day is:
[tex]E(Y)=\dfrac{550\cdot 20+150\cdot 40+80\cdot 60}{780}=\dfrac{21800}{2000}=27.95[/tex]
