Respuesta :
Answer:
The 95% confidence interval for the mean is lower confidence limit = 19.88, and upper confidence limit = 20.12. Some notations for confidence interval are (19.88, 20.12) or [19.88, 20.12].
Step-by-step explanation:
The answer to this question is to find two values that define an interval in which we can consider there is, with a certain probability, the population's mean [tex] \\ \mu[/tex].
We have here important information from the question:
- The size of the sample is "one thousand teens". So, n = 1000.
- The random variable "the number of hours they (the teens) spend on social networking sites each week" is normally distributed. In other words, this random variable follows a normal distribution.
- The value of the mean for this sample is [tex] \\ \overline{x} =20[/tex].
- The population standard deviation [tex] \\ \sigma =2\;hours[/tex].
- For a 95% confidence interval, we have a confidence coefficient of 1.96. The explanation for the latter is, roughly speaking, that we have to both side we have a remaining of 5%/2 or 0.05/2 = 0.025. The z-score corresponding to this probability is 1.96 (or -1.96, below the mean, and 1.96 above the mean. That is, [tex] \\ P(z< -1.96) = 0.025[/tex] and [tex] \\ P(z>1.96) = 0.025[/tex]).
Since we know that the random variable is normally distributed, the sample means are also normally distributed. Then, we can pose mathematically this problem as follows:
[tex] \\ P(\overline{x} - 1.96\frac{\sigma}{\sqrt{n}} \leq \mu \leq \overline{x} + 1.96\frac{\sigma}{\sqrt{n}}) = 0.95 [/tex] [1]
In this way, we have enough information to solve it:
[tex] \\ \overline{x} = 20[/tex].
[tex] \\ \sigma = 2[/tex].
[tex] \\ n = 1000[/tex].
Well, substituting each value in [1], we have:
[tex] \\ P(20 - 1.96\frac{2}{\sqrt{1000}} \leq \mu \leq 20 + 1.96\frac{2}{\sqrt{1000}}) = 0.95[/tex]
Then
[tex] \\ P(20 - 1.96\frac{2}{31.62} \leq \mu \leq 20 + 1.96\frac{2}{31.62}) = 0.95[/tex]
The value for
[tex] \\ 1.96\frac{2}{31.62} \approx 1.96*0.06 \approx 0.12[/tex]
Therefore
[tex] \\ P(20 - 0.12 \leq \mu \leq 20 + 0.12) = 0.95[/tex]
[tex] \\ P(19.88 \leq \mu \leq 20.12) = 0.95[/tex]
As a result, we can say that the 95% confidence interval for the mean is lower confidence limit = 19.88 and upper confidence limit = 20.12.
Note: some authors described these limits as (19.88, 20.12), while others as [19.88, 20.12].
