Question

1.17 A study of the effects of smoking on sleep patterns is conducted. The measure observed is the time, in minutes, that it takes to fall asleep. These data are obtained: Smokers: 69.3 56.0 22.1 47.6 53.2 48.1 52.7 34.4 60.2 43.8 23.2 13.8 Nonsmokers: 28.6 25.1 26.4 34.9 29.8 28.4 38.5 30.2 30.6 31.8 41.6 21.1 36.0 37.9 13.9 (a) Find the sample mean for each group. (b) Find the sample standard deviation for each group. (c) Make a dot plot of the data sets A and B on the same line. (d) Comment on what kind of impact smoking appears to have on the time required to fall asleep.

Answer 1

Answer:

a.

$\bar X_F=43.7$

$\bar X_{NF}=30.32$

b.

$S_F=16.9278$

$S_{NF}=7.12783$

c.

Attached file

d.

Apparently the practice of smoking reduces the ability to fall asleep, demanding much more time in individuals who smoke, than in those who do not smoke.

Step-by-step explanation:

a, b) For the group of smoking individuals, the average time it takes to fall asleep and the standard deviation of those times is:

$\bar X_F={\frac{1}{n} \sum_{i=1}^n x_i = 43.7$

$S_F=\sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i-\bar{x})^2}=16.9278$

a, b) For the group of non-smoking individuals, the average time it takes to fall asleep and the standard deviation of those times is:

$\bar X_{NF}={\frac{1}{n} \sum_{i=1}^n x_i = 30.32$

$S_{NF}=\sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i-\bar{x})^2}=7.12783$

c. In the attached file you can see the diagram of points for the times, in the smoking and non-smoking groups.

d. Apparently the practice of smoking reduces the ability to fall asleep, demanding much more time in individuals who smoke, than in those who do not smoke.

Answer 2

Smoking increases the time required for falling  asleep. The sample mean and standard deviation of the considered data are obtained as:

• Set A: Sample mean= 43.7 approx, Standard deviation = 16.93 approx
• Set B: Sample mean= 30.32 approx, Standard deviation = 7.13 approx

How to find the mean (expectation) and standard deviation of a data set?

Suppose that there are n (finite value) data observations available, then we get:

$Mean = \overline{x} = \dfrac{\sum_{\forall x_i}x_i}{n}\\\\Variance = \sigma^2 = \dfrac{\sum_{\forall x_i} (x_i-\overline{x})^2}{n}$

As standard deviation is positive root of variance, thus,

$\sigma = \sqrt{ \dfrac{\sum_{\forall x_i} (x_i-\overline{x})^2}{n}}$

where $x_i; \: \: i = 1,2, ... ,n$ is its n data values

Using the above formulas, we get:

For set A: There are n = 12 values. Their sum is 524.4

thus, $\overline{x} = \dfrac{524.4}{12} \approx 43.7$ = mean of the data set

Therefore, we get:

$\sigma = \sqrt{ \dfrac{\sum_{\forall x_i} (x_i-\overline{x})^2}{n}} = \sqrt{ \dfrac{\sum_{\forall x_i} (x_i-43.7)^2}{12}}$

Putting all data values, and evaluating it provides us standard deviation approximately 16.93

For set B: There are n = 15 values. Their sum is 454/8

thus, $\overline{x} = \dfrac{454.8}{15} \approx 30.32$ = mean of the data set

Therefore, we get:

$\sigma = \sqrt{ \dfrac{\sum_{\forall x_i} (x_i-\overline{x})^2}{n}} = \sqrt{ \dfrac{\sum_{\forall x_i} (x_i-30.32)^2}{15}}$

Putting all data values, and evaluating it provides us standard deviation approximately 7.13

Since mean of a data set is representative of its data values, and here, the time data values are denoting the time taken in minutes to get sleep, we see that data set A (of smokers) have higher mean than those who don't smoke. That means smoking increases the time required to fall sleep, therefore is bad for sleep.

Thus, the sample mean and standard deviation of the considered data are obtained as:

• Set A: Sample mean= 43.7 approx, Standard deviation = 16.93 approx
• Set B: Sample mean= 30.32 approx, Standard deviation = 7.13 approx

Smokers require more time for falling  asleep than those who do not smoke.

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