Respuesta :
Answer:
a) The 95% confidence interval estimate of the population mean is between 6.206 hours and 6.794 hours.
b) The confidence interval means that we are 95% sure that the true population mean for the amount of time teenagers spend on the internet is between 6.206 hours and 6.794 hours.
Step-by-step explanation:
a)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].
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.
[tex]M = 1.96*\frac{1.5}{\sqrt{100}} = 0.294[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 6.5 - 0.294 = 6.206 hours.
The upper end of the interval is the sample mean added to M. So it is 6.5 + 0.294 = 6.794 hours.
The 95% confidence interval estimate of the population mean is between 6.206 hours and 6.794 hours.
b)Interpretation:
The confidence interval means that we are 95% sure that the true population mean for the amount of time teenagers spend on the internet is between 6.206 hours and 6.794 hours.
Following are the solution to the given question:
- In the given scenario, we assume that the true population will be [tex]\mu[/tex]. At [tex]5\%[/tex] level of significance, it cuts off the value that [tex]\pm 1.96[/tex].
[tex]\to -1.96 \leq \frac{X-\mu}{sd_x} \times \sqrt{n} \leq 1.96[/tex]
[tex]\to X = 6.5 \ \ and\ \ sd_x = 1.5 \\\\\to -1.96 \leq \frac{6.5 - \mu}{1.5} \times \sqrt{100} \leq 1.96\\\\ \to -1.96 \leq \frac{6.5 - \mu}{1.5} \times 10 \leq 1.96\\\\ \to -.294 \leq 6.5 - \mu \leq 0.294 \\\\ \to -6.794 \leq - \mu \leq -6.206[/tex]
- A population parameter is estimated with a 95% confidence interval.
- It suggests that if 100 similar intervals are established, 95% of the genuine population mean will fall inside this interval.
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