Human skin loses water through evaporation, and, in general, damaged skin loses water at a faster rate than undamaged skin. In a study in which the outer layer of skin on a small section of the forearm was partially removed, water loss (in g/m^2 per hour) was measured both before and after skin removal. The results for ten individuals were as follows:

Question

subject 1- before: 13 , after: 20
2- 13/19
3- 14/28
4- 11/27
5- 15/25
6- 17/26
7- 17/25
8- 18/23
9- 17/21
10- 14/20

Construct a 98% confidence interval. Let d represent water loss after skin removal minute the water loss before skin removal. Round to 1 decimal place

Answer 1

To construct a 98% confidence interval for the difference in water loss before and after skin removal (d), we first need to calculate the differences for each subject, find the mean and standard deviation of these differences, and then use these values to compute the confidence interval.

Step 1: Calculate the differences (d)

We'll compute \( d \) for each subject as follows: \[ d = \text{water loss after} - \text{water loss before} \]

| Subject | Before (g/m²/h) | After (g/m²/h) | Difference (d) | |---------|-----------------|----------------|-----------------| | 1 | 13 | 20 | 20 - 13 = 7 | | 2 | 13 | 19 | 19 - 13 = 6 | | 3 | 14 | 28 | 28 - 14 = 14 | | 4 | 11 | 27 | 27 - 11 = 16 | | 5 | 15 | 25 | 25 - 15 = 10 | | 6 | 17 | 26 | 26 - 17 = 9 | | 7 | 17 | 25 | 25 - 17 = 8 | | 8 | 18 | 23 | 23 - 18 = 5 | | 9 | 17 | 21 | 21 - 17 = 4 | | 10 | 14 | 20 | 20 - 14 = 6 |

Now let's present the differences: \[ d = [7, 6, 14, 16, 10, 9, 8, 5, 4, 6] \]

Step 2: Calculate the mean and standard deviation of the differences

Mean of d

\[ \bar{d} = \frac{\sum d}{n} \] where \( n \) is the number of subjects: \[ \sum d = 7 + 6 + 14 + 16 + 10 + 9 + 8 + 5 + 4 + 6 = 85 \] \[ n = 10 \] \[ \bar{d} = \frac{85}{10} = 8.5 \]

Standard deviation of d

First, we compute the squared differences from the mean: \[ \text{Variance} = s^2 = \frac{\sum (d_i - \bar{d})^2}{n-1} \] Calculating each squared difference:

• For 7: \( (7 - 8.5)^2 = 2.25 \)
• For 6: \( (6 - 8.5)^2 = 6.25 \)
• For 14: \( (14 - 8.5)^2 = 30.25 \)
• For 16: \( (16 - 8.5)^2 = 56.25 \)
• For 10: \( (10 - 8.5)^2 = 2.25 \)
• For 9: \( (9 - 8.5)^2 = 0.25 \)
• For 8: \( (8 - 8.5)^2 = 0.25 \)
• For 5: \( (5 - 8.5)^2 = 12.25 \)
• For 4: \( (4 - 8.5)^2 = 20.25 \)
• For 6: \( (6 - 8.5)^2 = 6.25 \)

Summing these values: \[ \sum (d_i - \bar{d})^2 = 2.25 + 6.25 + 30.25 + 56.25 + 2.25 + 0.25 + 0.25 + 12.25 + 20.25 + 6.25 = 142.5 \] Thus, the variance is: \[ s^2 = \frac{142.5}{10 - 1} = \frac{142.5}{9} \approx 15.8333 \] Standard deviation: \[ s \approx \sqrt{15.8333} \approx 3.98 \]

Step 3: Construct the confidence interval

To find the 98% confidence interval, we will use the t-distribution since the sample size is small. The degrees of freedom \( df = n - 1 = 9 \).

Looking up the t-value for \( df = 9 \) at a 98% confidence level, we find \( t_{0.01, 9} \approx 2.821 \).

The margin of error (ME) is given by: \[ \text{ME} = t \times \frac{s}{\sqrt{n}} = 2.821 \times \frac{3.98}{\sqrt{10}} \approx 2.821 \times 1.258 \approx 3.55 \]

Now we can calculate the confidence interval: \[ \bar{d} - \text{ME} \quad \text{to} \quad \bar{d} + \text{ME} \] \[ 8.5 - 3.55 \quad \text{to} \quad 8.5 + 3.55 \] \[ 4.95 \quad \text{to} \quad 12.05 \]

Therefore, the 98% confidence interval for the difference in water loss is approximately: \[ \boxed{(5.0, 12.1)} \] (rounded to one decimal place).

Answer 2

(b)A dermatologist claims that the difference in water loss is 8 g/m^2 per hour. Does the confidence interval contradict this claim?

Answer 3

To determine whether the dermatologist's claim that the difference in water loss is 8 g/m² per hour is contradicted by the 98% confidence interval we calculated in part (a), we need to examine the confidence interval:

The 98% confidence interval for the difference in water loss was approximately:
\[ (5.0, 12.1) \]

Analysis of the Claim:

The claim states that the difference in water loss is exactly 8 g/m² per hour.

The confidence interval \( (5.0, 12.1) \) indicates that we are 98% confident that the true mean difference in water loss (after skin removal minus before skin removal) lies within this range.

Evaluation:

• The interval \( (5.0, 12.1) \) includes the value 8.
• Since 8 falls within the range of our confidence interval, this means that the data does not contradict the dermatologist's claim.

Conclusion:

The confidence interval does not contradict the claim that the difference in water loss is 8 g/m² per hour. In fact, it suggests that it is a plausible value given the observed data.