Respuesta :
Answer:
(a) 95% confidence interval for the true average CO2 level in the population of all homes is [607.42 , 700.90].
(b) Sample size that would be necessary to obtain an interval width of 50 ppm for a confidence level of 95% is at least 50.
Step-by-step explanation:
We are given that a sample of 50 kitchens with gas cooking appliances monitored during a one week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 164.43.
(a) Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean CO2 level = 654.16
s = sample standard deviation = 164.43
n = sample of kitchens = 50
[tex]\mu[/tex] = true average CO2 level
Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.01 < [tex]t_4_9[/tex] < 2.01) = 0.95 {As the critical value of t at 49 degree
of freedom are -2.01 & 2.01 with P = 2.5%}
P(-2.01 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.01) = 0.95
P( [tex]-2.01 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.01 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.01 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.01 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.01 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.01 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]654.16-2.01 \times {\frac{164.43}{\sqrt{50} } }[/tex] , [tex]654.16+2.01 \times {\frac{164.43}{\sqrt{50} } }[/tex] ]
= [607.42 , 700.90]
Therefore, 95% confidence interval for the true average CO2 level in the population of all homes is [607.42 , 700.90].
(b) We are given that the investigators had made a rough guess of 175 for the value of s before collecting data and from this we have to find the sample size that would be necessary to obtain an interval width of 50 ppm for a confidence level of 95%;
Here, we will use the concept of Margin of error as the statement "to obtain an interval width of 50 ppm" represents the margin of error we want.
SO, Margin of error formula is given by;
Margin of error = [tex]Z_(_\frac{\alpha}{2}_ ) \times \frac{s}{\sqrt{n} }[/tex]
where, [tex]\alpha[/tex] = significance level = 5%
s = standard deviation = 175
n = sample size
Now, in the t table the critical value of x at 2.5% ( [tex]\frac{0.05}{2} = 0.025[/tex] ) level of significance is 2.01.
SO, Margin of error = [tex]Z_(_\frac{\alpha}{2}_ ) \times \frac{s}{\sqrt{n} }[/tex]
50 ppm = [tex]2.01 \times \frac{175}{\sqrt{n} }[/tex]
[tex]\sqrt{n} =\frac{175\times 2.01 }{50}[/tex]
[tex]\sqrt{n} =7.035[/tex]
n = 49.49 ≈ 50
Therefore, sample size that would be necessary to obtain an interval width of 50 ppm for a confidence level of 95% is at least 50.
