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Answer:
The point estimate for the true difference between the population means is of -2.62 years.
The 95% confidence interval for the difference between the true mean ages for cars owned by students and faculty is between -3.4 years and -1.84 years.
Step-by-step explanation:
Before building the confidence interval, we need to understand the central limit theorem, and subtraction between normal variables.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Subtraction of normal variables:
When we subtract normal variables, the mean is the subtraction of the mean, while the standard deviation is the square root of the sum of variances.
A sample of 138 cars owned by students had an average age of 5.13 years. The population standard deviation for cars owned by students is 3.45 years.
This means that:
[tex]\mu_{s} = 5.13, \sigma_{s} = 3.45, n = 138, s_s = \frac{3.45}{\sqrt{138}} = 0.2937[/tex]
A sample of 111 cars owned by faculty had an average age of 7.75 years. The population standard deviation for cars owned by faculty is 2.08 years.
This means that [tex]\mu_{f} = 7.75, \sigma_{f} = 2.08, n = 111, s_f = \frac{2.08}{\sqrt{111}} = 0.2658[/tex]
Difference between the true mean ages for cars owned by students and faculty.
s - f
Mean:
[tex]\mu = \mu_s - \mu_f = 5.13 - 7.75 = -2.62[/tex]
This is the point estimate for the true difference between the population means.
Standard deviation:
[tex]s = \sqrt{s_s^2+s_f^2} = \sqrt{0.2937^2+0.2658^2} = 0.3961[/tex]
Confidence interval:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = zs = 1.96*0.3961 = 0.78[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is -2.62 - 0.78 = -3.4
The upper end of the interval is the sample mean added to M. So it is -2.62 + 0.78 = -1.84
The 95% confidence interval for the difference between the true mean ages for cars owned by students and faculty is between -3.4 years and -1.84 years.
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