Respuesta :
Answer:
A 95% confidence interval for the true mean is [$3.39, $6.01].
Step-by-step explanation:
We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;
Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.
Firstly, the pivotal quantity for finding the 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 income = [tex]\frac{\sum X}{n}[/tex] = $4.70
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = $1.83
n = sample of parking meters = 10
[tex]\mu[/tex] = population mean
Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.262 < [tex]t_9[/tex] < 2.262) = 0.95 {As the critical value of t at 9 degrees of
freedom are -2.262 & 2.262 with P = 2.5%}
P(-2.262 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.262) = 0.95
P( [tex]-2.262 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu[/tex] < [tex]2.262 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.262 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.262 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.262 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.262 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }[/tex] , [tex]4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }[/tex] ]
= [$3.39, $6.01]
Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].
The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.
Also, the margin of error = [tex]2.262 \times {\frac{s}{\sqrt{n} } }[/tex]
= [tex]2.262 \times {\frac{1.83}{\sqrt{10} } }[/tex] = 1.31
Using the t-distribution, it is found that:
a) The 95% confidence interval for the true mean is (3.39, 6.01). It means that we are 95% sure that the true mean income for all parking meters in the resort community from which the sample was taken is between these two values.
b) The margin of error is of $1.31.
Item a:
We will have the standard deviation for the sample, which is why the t-distribution is used to solve this question.
The sample size given is of [tex]n = 10[/tex], and using a calculator, it is found that:
The sample mean of [tex]\overline{x} = 4.7[/tex].
The sample standard deviation of [tex]s = 1.833[/tex].
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 10 - 1 = 9 df, is t = 2.2622.
Then, the interval is:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 4.7 - 2.2622\frac{1.833}{\sqrt{10}} = 3.39[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 4.7 + 2.2622\frac{1.833}{\sqrt{10}} = 6.01[/tex]
The 95% confidence interval for the true mean is (3.39, 6.01). It means that we are 95% sure that the true mean income for all parking meters in the resort community from which the sample was taken is between these two values.
Item b:
The margin of error is half the distance between the two bounds, hence:
[tex]M = \frac{6.01 - 3.39}{2} = 1.31[/tex]
A similar problem is given at https://brainly.com/question/22596713
