Respuesta :
Answer:
The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.
Step-by-step explanation:
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].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
The standard deviation is the square root of the variance. So
[tex]\sigma = \sqrt{0.16} = 0.4[/tex]
Then
[tex]M = 1.96*\frac{0.4}{\sqrt{400}} = 0.0392[/tex]
The lower end of the interval is the mean subtracted by M. So it is 1.75 - 0.0392 = 1.7108m
The upper end of the interval is the mean added to M. So it is 1.75 + 0.0392 = 1.7892m
The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.
Answer:
- The interval of 95% confidence for the average heights of the population is = [tex](1.7108, 1.7892)[/tex]
Step-by-step explanation:
mean x = [tex]1.75[/tex]
Variance [tex]\rho^2 = 0.16[/tex]
standard deviation [tex](\rho) = \sqrt{0.16} = 0.4[/tex]
n = 400
[tex]95\%[/tex] confidence :
[tex]\alpha = 100\% - 95\% = 5\%\\\\\frac{\alpha}{2} = 2.5\% = 0.025[/tex]
From standard normal distribution table,
[tex]Z_\frac{\alpha}{2} = Z_{0.025} = 1.96[/tex]
Margin of error, [tex]E = Z_\frac{\alpha}{2} * \frac{\rho}{\sqrt{n}}[/tex]
[tex]E = 1.96 * \frac{0.4}{\sqrt{400}}\\\\E = 0.0392[/tex]
Lower limit: x - E
[tex]= 1.75 - 0.0392\\\\= 1.7108[/tex]
Upper limit: x + E
[tex]= 1.75 + 0.0392\\\\= 1.7892[/tex]
[tex]Limits : (1.7108, 1.7892)[/tex]
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