Respuesta :
Answer:
The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 pounds.
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 the margin of error 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.
[tex]M = 1.96*\frac{1.4}{\sqrt{2092}} = 0.1[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 2.9 - 0.1 = 2.8 pounds
The upper end of the interval is the sample mean added to M. So it is 2.9 + 0.1 = 3 pounds.
The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 pounds.
Answer:
[tex]2.9-1.96\frac{1.4}{\sqrt{2092}}=2.84[/tex]
[tex]2.9+1.96\frac{1.4}{\sqrt{2092}}=2.96[/tex]
The confidence interval is given by [tex]2.84 \leq \mu \leq 2.96[/tex]
Step-by-step explanation:
Information given
[tex]\bar X=2.9[/tex] represent the sample mean
[tex]\mu[/tex] population mean
[tex]\sigma =1.4[/tex] represent the population standard deviation
n=2092 represent the sample size
Confidence interval
The confidence interval is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence interval is 0.95 or 95%, the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value would be [tex]z_{\alpha/2}=1.96[/tex]
Replacing the info we got:
[tex]2.9-1.96\frac{1.4}{\sqrt{2092}}=2.84[/tex]
[tex]2.9+1.96\frac{1.4}{\sqrt{2092}}=2.96[/tex]
The confidence interval is given by [tex]2.84 \leq \mu \leq 2.96[/tex]
