A simple random sample of eight college freshmen were asked how many hours of sleep they typically got per night. The results were

Question

8.5
7.5
6.5
9
6
24
8.5
8

(a) The data contain an outlier that is clearly a mistake. Eliminate the outlier, then construct a 90% confidence interval for the mean amount of sleep. Round the answers to at least two decimal places.

Answer 1

To identify and eliminate the outlier from the data, we can observe that the value "24" is significantly higher than the other values. Hence, we will remove it from the dataset.

The corrected data set after removing the outlier is:

• 8.5
• 7.5
• 6.5
• 9
• 6
• 8.5
• 8

Step 1: Calculate the Mean and Standard Deviation

1. Calculate the mean ($\overline{x}$): $$\overline{x}=\frac{(8.5+7.5+6.5+9+6+8.5+8)}{7}=\frac{54.5}{7}\approx 7.7857$$

2. Calculate the sample standard deviation (s): $$s=\sqrt{\frac{1}{n-1}\sum (x_i-\overline{x}{)}^2}$$ $$=\sqrt{\frac{1}{6}((8.5-7.7857{)}^2+(7.5-7.7857{)}^2+(6.5-7.7857{)}^2+(9-7.7857{)}^2+(6-7.7857{)}^2+(8.5-7.7857{)}^2+(8-7.7857{)}^2)}$$

Calculating the squared differences:

• $(8.5-7.7857{)}^2\approx 0.5102$
• $(7.5-7.7857{)}^2\approx 0.0802$
• $(6.5-7.7857{)}^2\approx 1.6510$
• $(9-7.7857{)}^2\approx 1.4596$
• $(6-7.7857{)}^2\approx 3.1623$
• $(8.5-7.7857{)}^2\approx 0.5102$
• $(8-7.7857{)}^2\approx 0.0460$

Now sum these squared differences: $$=0.5102+0.0802+1.6510+1.4596+3.1623+0.5102+0.0460\approx 7.4195$$

Now plug into the standard deviation formula: $$s=\sqrt{\frac{7.4195}{6}}\approx \sqrt{1.2366}\approx 1.11$$

Step 2: Construct the Confidence Interval

We'll use the t-distribution to calculate the 90% confidence interval since the sample size is small ($n=7$).

1. Degrees of freedom: $df=n-1=6$

2. Find the t-value for a 90% confidence interval: Using a t-table, the t-score for $df=6$ at 90% confidence is approximately $t_{0.05,6}\approx 1.943$.

3. Calculate the margin of error (ME): $$ME=t\times \frac{s}{\sqrt{n}}=1.943\times \frac{1.11}{\sqrt{7}}\approx 1.943\times \frac{1.11}{2.6458}\approx 1.943\times 0.419\approx 0.8134$$

4. Construct the confidence interval: $$\text{Lower Limit}=\overline{x}-ME\approx 7.7857-0.8134\approx 6.9723$$ $$\text{Upper Limit}=\overline{x}+ME\approx 7.7857+0.8134\approx 8.5991$$

Step 3: Round to Two Decimal Places

Thus, the 90% confidence interval for the mean amount of sleep is approximately: $$\boxed{{\displaystyle (6.97,8.60)}}$$

Answer 2

(b) Leave the outlier in and construct the 90% confidence interval. Round the answers to at least two decimal places.

Answer 3

To construct the 90% confidence interval for the mean amount of sleep including the outlier (24), we'll follow the same steps as before but use the full dataset.

Full Dataset

The dataset including the outlier is:

• 8.5
• 7.5
• 6.5
• 9
• 6
• 24
• 8.5
• 8

Step 1: Calculate the Mean and Standard Deviation

1. Calculate the mean ($\overline{x}$): $$\overline{x}=\frac{(8.5+7.5+6.5+9+6+24+8.5+8)}{8}=\frac{70}{8}=8.75$$

2. Calculate the sample standard deviation (s): We first calculate the squared differences from the mean:

   $$s=\sqrt{\frac{1}{n-1}\sum (x_i-\overline{x}{)}^2}$$

Calculating the squared differences:

• $(8.5-8.75{)}^2\approx 0.0625$
• $(7.5-8.75{)}^2\approx 1.5625$
• $(6.5-8.75{)}^2\approx 5.0625$
• $(9-8.75{)}^2\approx 0.0625$
• $(6-8.75{)}^2\approx 7.5625$
• $(24-8.75{)}^2\approx 234.0625$
• $(8.5-8.75{)}^2\approx 0.0625$
• $(8-8.75{)}^2\approx 0.5625$

Now sum these squared differences: $$=0.0625+1.5625+5.0625+0.0625+7.5625+234.0625+0.0625+0.5625\approx 248.5$$

Now plug into the standard deviation formula: $$s=\sqrt{\frac{248.5}{7}}\approx \sqrt{35.5}\approx 5.96$$

Step 2: Construct the Confidence Interval

We'll use the t-distribution to calculate the 90% confidence interval since the sample size is small ($n=8$).

1. Degrees of freedom: $df=n-1=7$

2. Find the t-value for a 90% confidence interval: Using a t-table, the t-score for $df=7$ at 90% confidence is approximately $t_{0.05,7}\approx 1.895$.

3. Calculate the margin of error (ME): $$ME=t\times \frac{s}{\sqrt{n}}=1.895\times \frac{5.96}{\sqrt{8}}\approx 1.895\times \frac{5.96}{2.828}\approx 1.895\times 2.11\approx 4.002$$

4. Construct the confidence interval: $$\text{Lower Limit}=\overline{x}-ME\approx 8.75-4.002\approx 4.748$$ $$\text{Upper Limit}=\overline{x}+ME\approx 8.75+4.002\approx 12.752$$

Step 3: Round to Two Decimal Places

Thus, the 90% confidence interval for the mean amount of sleep including the outlier is approximately: $$\boxed{{\displaystyle (4.75,12.75)}}$$