Respuesta :
Answer:
a) 16.6
b) The 95% confidence level of the mean of the time playing video games is between 15.7 and 17.5 hours.
c) The 99% confidence level of the mean of the time playing video games is between 15.4 and 17.8 hours.
d) The margin of error increases as the confidence level increases, due to the value of z, which means that the 99% confidence interval is larger.
Step-by-step explanation:
a. Find the best point of estimate of the population mean
The best estimate for the population mean is the sample mean, which is 16.6
b. Find the 95% confidence level of the mean of the time playing video games
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 = 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{2.8}{\sqrt{35}} = 0.9[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 16.6 - 0.9 = 15.7 hours
The upper end of the interval is the sample mean added to M. So it is 16.6 - 0.9 = 17.5 hours
The 95% confidence level of the mean of the time playing video games is between 15.7 and 17.5 hours.
c. Find the 99% confidence interval of the mean time playing video games
Following the same logic as above, we have that [tex]Z = 2.575[/tex]. So
[tex]M = 2.575\frac{2.8}{\sqrt{35}} = 0.9[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 16.6 - 1.2 = 15.4 hours.
The upper end of the interval is the sample mean added to M. So it is 16.6 + 1.2 = 17.8 hours.
The 99% confidence level of the mean of the time playing video games is between 15.4 and 17.8 hours.
d. Which is larger? Explain why.
The margin of error increases as the confidence level increases, due to the value of z, which means that the 99% confidence interval is larger.
The mean is 16.6, 95% confidence interval is between 15.68 to 17.52, the 99% confidence interval is between 15.39 to 17.81, and the 99% confidence interval is larger.
It is given that the in recent study of 35 ninth-grade students the mean number of hours per week is 16.6 with standard deviation of the population was 2.8.
It is required to find the best point of estimate of the population mean, 95% confidence interval, and 99% confidence interval.
What is a confidence interval for population standard deviation?
It is defined as the sampling distribution following an approximately normal distribution for known standard deviation.
We have confidence level = 95% = 0.95
We know:
[tex]\alpha = \frac{1-0.95}{2}[/tex] = 0.025
[tex]\rm Z_1 _-_\alpha = Z_{0.975} = 1.96[/tex] (from the Z table)
The sample size n = 35
We know the Margin of error formula:
[tex]\rm M =Z\frac{\sigma}{\sqrt[]{n} }[/tex] ( [tex]\rm \sigma\\[/tex] = 2.8 )
[tex]\rm M =1.96\frac{2.8}{\sqrt[]{35} }[/tex]
M = 0.92
The 95% interval will be:
= X - M ⇒ 16.6 - 0.92 ⇒15.68
= X+ M ⇒ 16.6+0.92 ⇒ 17.52
For 99% confidence level:
[tex]\alpha = \frac{1-0.90}{2} = 0.005[/tex]
[tex]\rm Z_1 _-_\alpha = Z_{0.005} = 2.575[/tex] (from Z table)
[tex]\rm M =2.575\frac{2.8}{\sqrt[]{35} }[/tex]
M = 1.21
The 99% interval will be:
=16.6 - 1.21 ⇒ 15.39
= 16.6+ 1.21 = 17.81
As the value of M increases the confidence value increases due to the value of Z that means 99% confidence interval is larger.
Thus, the mean is 16.6, 95% confidence interval is between 15.68 to 17.52, the 99% confidence interval is between 15.39 to 17.81, and the 99% confidence interval is larger.
Learn more about the confidence interval here:
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