Thank you for visiting Sleeping Outlier Analysis A simple random sample of nine college freshmen were asked how many hours of sleep they typically got per night The results. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
We start with the original data:
[tex]$$
9.5,\ 8.5,\ 24,\ 8,\ 9,\ 6.5,\ 7.5,\ 6,\ 8.5,
$$[/tex]
and notice that the value [tex]$24$[/tex] is clearly a mistake. Removing this outlier, the cleaned data becomes:
[tex]$$
9.5,\ 8.5,\ 8,\ 9,\ 6.5,\ 7.5,\ 6,\ 8.5.
$$[/tex]
Since there are [tex]$8$[/tex] values now, the sample size is [tex]$n = 8$[/tex].
1. Calculate the sample mean:
The sample mean [tex]$\overline{x}$[/tex] is given by
[tex]$$
\overline{x} = \frac{9.5 + 8.5 + 8 + 9 + 6.5 + 7.5 + 6 + 8.5}{8} = 7.9375.
$$[/tex]
2. Determine the sample standard deviation:
With the cleaned data, using Bessel's correction (dividing by [tex]$n-1$[/tex]), suppose we find
[tex]$$
s \approx 1.20823.
$$[/tex]
3. Find the critical [tex]$t$[/tex]-value:
We are to construct a [tex]$99.8\%$[/tex] confidence interval. This corresponds to a significance level [tex]$\alpha = 1 - 0.998 = 0.002$[/tex], so that each tail has [tex]$\alpha/2 = 0.001$[/tex]. With [tex]$n-1 = 7$[/tex] degrees of freedom, the corresponding critical [tex]$t$[/tex]-value is approximately
[tex]$$
t^* \approx 4.78529.
$$[/tex]
4. Compute the standard error of the mean (SEM):
The standard error is calculated as
[tex]$$
SE = \frac{s}{\sqrt{n}} \approx \frac{1.20823}{\sqrt{8}}.
$$[/tex]
5. Calculate the margin of error:
The margin of error is
[tex]$$
\text{Margin} = t^* \times SE \approx 4.78529 \times \frac{1.20823}{\sqrt{8}} \approx 2.04415.
$$[/tex]
6. Construct the confidence interval:
The [tex]$99.8\%$[/tex] confidence interval for the mean is
[tex]$$
\overline{x} \pm \text{Margin} \quad \Rightarrow \quad 7.9375 \pm 2.04415.
$$[/tex]
This gives:
[tex]$$
\text{Lower bound} = 7.9375 - 2.04415 \approx 5.89335, \quad \text{Upper bound} = 7.9375 + 2.04415 \approx 9.98165.
$$[/tex]
7. Round to two decimal places:
[tex]$$
5.89 < \mu < 9.98.
$$[/tex]
Thus, the [tex]$99.8\%$[/tex] confidence interval for the mean amount of sleep from the cleaned data is
[tex]$$
5.89 < \mu < 9.98.
$$[/tex]
[tex]$$
9.5,\ 8.5,\ 24,\ 8,\ 9,\ 6.5,\ 7.5,\ 6,\ 8.5,
$$[/tex]
and notice that the value [tex]$24$[/tex] is clearly a mistake. Removing this outlier, the cleaned data becomes:
[tex]$$
9.5,\ 8.5,\ 8,\ 9,\ 6.5,\ 7.5,\ 6,\ 8.5.
$$[/tex]
Since there are [tex]$8$[/tex] values now, the sample size is [tex]$n = 8$[/tex].
1. Calculate the sample mean:
The sample mean [tex]$\overline{x}$[/tex] is given by
[tex]$$
\overline{x} = \frac{9.5 + 8.5 + 8 + 9 + 6.5 + 7.5 + 6 + 8.5}{8} = 7.9375.
$$[/tex]
2. Determine the sample standard deviation:
With the cleaned data, using Bessel's correction (dividing by [tex]$n-1$[/tex]), suppose we find
[tex]$$
s \approx 1.20823.
$$[/tex]
3. Find the critical [tex]$t$[/tex]-value:
We are to construct a [tex]$99.8\%$[/tex] confidence interval. This corresponds to a significance level [tex]$\alpha = 1 - 0.998 = 0.002$[/tex], so that each tail has [tex]$\alpha/2 = 0.001$[/tex]. With [tex]$n-1 = 7$[/tex] degrees of freedom, the corresponding critical [tex]$t$[/tex]-value is approximately
[tex]$$
t^* \approx 4.78529.
$$[/tex]
4. Compute the standard error of the mean (SEM):
The standard error is calculated as
[tex]$$
SE = \frac{s}{\sqrt{n}} \approx \frac{1.20823}{\sqrt{8}}.
$$[/tex]
5. Calculate the margin of error:
The margin of error is
[tex]$$
\text{Margin} = t^* \times SE \approx 4.78529 \times \frac{1.20823}{\sqrt{8}} \approx 2.04415.
$$[/tex]
6. Construct the confidence interval:
The [tex]$99.8\%$[/tex] confidence interval for the mean is
[tex]$$
\overline{x} \pm \text{Margin} \quad \Rightarrow \quad 7.9375 \pm 2.04415.
$$[/tex]
This gives:
[tex]$$
\text{Lower bound} = 7.9375 - 2.04415 \approx 5.89335, \quad \text{Upper bound} = 7.9375 + 2.04415 \approx 9.98165.
$$[/tex]
7. Round to two decimal places:
[tex]$$
5.89 < \mu < 9.98.
$$[/tex]
Thus, the [tex]$99.8\%$[/tex] confidence interval for the mean amount of sleep from the cleaned data is
[tex]$$
5.89 < \mu < 9.98.
$$[/tex]
Thank you for reading the article Sleeping Outlier Analysis A simple random sample of nine college freshmen were asked how many hours of sleep they typically got per night The results. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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Rewritten by : Jeany
