Answer:
The 68% confidence interval is (6.3, 6.7).
The 95% confidence interval is (6.1, 6.9).
The 99.7% confidence interval is (5.9, 7.1).
Step-by-step explanation:
The Central Limit Theorem states that if we have a population with mean μ and standard deviation σ and take appropriately huge random-samples (n ≥ 30) from the population with replacement, then the distribution of the sample-means will be approximately normally distributed.
Then, the mean of the sample means is given by,
[tex]\mu_{\bar x}=\bar x[/tex]
And the standard deviation of the sample means (also known as the standard error)is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}} \ \text{or}\ \frac{s}{\sqrt{n}}[/tex]
The information provided is:
[tex]n=400\\\\\bar x=6.5\\\\s=4[/tex]
As n = 400 > 30, the sampling distribution of the sample-means will be approximately normally distributed.
(a)
Compute the 68% confidence interval for population mean as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\cdot \frac{s}{\sqrt{n}}[/tex]
[tex]=6.5\pm 0.9945\cdot \frac{4}{\sqrt{400}}\\\\=6.5\pm 0.1989\\\\=(6.3011, 6.6989)\\\\\approx (6.3, 6.7)[/tex]
The 68% confidence interval is (6.3, 6.7).
The margin of error is:
[tex]MOE=\frac{UL-LL}{2}=\frac{6.7-6.3}{2}=0.20[/tex]
(b)
Compute the 95% confidence interval for population mean as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\cdot \frac{s}{\sqrt{n}}[/tex]
[tex]=6.5\pm 1.96\cdot \frac{4}{\sqrt{400}}\\\\=6.5\pm 0.392\\\\=(6.108, 6.892)\\\\\approx (6.1, 6.9)[/tex]
The 95% confidence interval is (6.1, 6.9).
The margin of error is:
[tex]MOE=\frac{UL-LL}{2}=\frac{6.9-6.1}{2}=0.40[/tex]
(c)
Compute the 99.7% confidence interval for population mean as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\cdot \frac{s}{\sqrt{n}}[/tex]
[tex]=6.5\pm 0.594\cdot \frac{4}{\sqrt{400}}\\\\=6.5\pm 0.392\\\\=(5.906, 7.094)\\\\\approx (5.9, 7.1)[/tex]
The 99.7% confidence interval is (5.9, 7.1).
The margin of error is:
[tex]MOE=\frac{UL-LL}{2}=\frac{7.1-5.9}{2}=0.55[/tex]
