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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]
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