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Laptop Computers and Sperm Count Stu dies have shown that heating the scrotum by jus \(1^{\circ} \mathrm{C}\) can reduce sperm count and sperm quality so men concerned about fertility are cautioned to avoid too much time in the hot tub or sauna. A new study \(^{41}\) suggests that men also keep their lap top computers off their laps. The study measurec scrotal temperature in 29 healthy male volunteer as they sat with legs together and a laptop compute on the lap. Temperature increase in the left scrotun over a 60 -minute session is given as \(2.31 \pm 0.96\) anc a note tells us that "Temperatures are given as \({ }^{\circ} \mathrm{C}\) values are shown as mean \(\pm \mathrm{SD} . "\) The abbreviatior SD stands for standard deviation. (Men who sit witl their legs together without a laptop computer do not show an increase in temperature.) (a) If we assume that the distribution of the temper ature increases for the 29 men is symmetric anc bell-shaped, find an interval that we expect to contain about \(95 \%\) of the temperature increases (b) Find and interpret the \(z\) -score for one of the men, who had a temperature increase of \(4.9^{\circ}\).

Step by step solution

01

Understand the given data and find the interval

We are given that the average increase in scrotal temperature is \(2.31^{\circ}\) with a standard deviation of \(0.96^{\circ}\). Since we expect approximately \(95\%\) of observations to fall within 2 standard deviations of the mean in a normal distribution, we can calculate our expected interval by subtracting and adding 2 standard deviations from the mean. This would be \(2.31 - 2*0.96\) and \(2.31 + 2*0.96\).

02

Calculate the lower and upper bounds of the interval

The computation gives us a lower bound of \(2.31 - 2*0.96 = 0.39^{\circ}\) and an upper bound of \(2.31 + 2*0.96 = 4.23^{\circ}\). This is the range we expect \(95\%\) of temperature increases to fall within.

03

Interpret the z-score

The z-score for one of the men with a temperature increase of \(4.9^{\circ}\) can be calculated by subtracting the mean from the observed value and dividing by the standard deviation. This gives us \(z = (4.9 - 2.31) / 0.96\).

04

Compute the z-score

This calculation gives us a z-score of \(z = (4.9 - 2.31) / 0.96 = 2.7\) approximately.

05

Interpret the z-score

A z-score of 2.7 indicates that the observed temperature increase of \(4.9^{\circ}\) is 2.7 standard deviations above the mean. This means this temperature increase is significantly higher than the average increase for the group of men studied in this exercise. Because it is more than two standard deviations from the mean, we can conclude that it is a relatively rare event in a normal distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation

In statistics, the standard deviation is a measure of how spread out numbers are in a data set. In simpler terms, it tells us the average distance from the mean for each data point.

For example, if we take a look at the exercise, the average temperature increase observed is 2.31°C. The standard deviation, 0.96°C, indicates the typical amount by which the individual temperature increases deviate from this average.

• The smaller the standard deviation, the closer the temperatures are to the average (mean).
• A larger standard deviation indicates more variation in the temperature increases.

Understanding standard deviation is crucial because it provides a clear picture of the variability within the data.

In a normal distribution, such as the one assumed in the problem, about 68% of the data falls within one standard deviation of the mean, and 95% falls within two standard deviations. This property helps us calculate expected ranges for data.

Z-Score

The z-score is a statistical measure that describes how many standard deviations a particular data point is from the mean. It helps in understanding the position of that data point within the distribution.

A z-score is calculated with the formula: \[ z = \frac{(X - \mu)}{\sigma} \]where

• \(X\) is your data point,
• \(\mu\) is the mean of the data set,
• and \(\sigma\) is the standard deviation.

In the exercise, for a temperature increase of 4.9°C, we calculate the z-score: \[ z = \frac{(4.9 - 2.31)}{0.96} \approx 2.7 \]
This z-score tells us that the 4.9°C is 2.7 standard deviations above the mean, indicating it is much higher than the average increase and suggests this temperature measurement is somewhat unusual, given the normal distribution of data assumed in this problem.

• Z-scores help determine if a data point is common or rare in a given context.

Mean

The mean, often referred to as the average, is one of the most common descriptive statistics. It provides a single value representing a central tendency of the data set. To find the mean, you sum up all the data points and divide by the number of points.

In the laptop and sperm count study, the mean temperature increase was given as 2.31°C. This represents the typical or average increase in temperature when participants sat with a laptop on their laps for the study's duration.

• The mean is useful for giving a quick snapshot of the data set's typical value.
• It is sensitive to outliers, and this is why the standard deviation is also important: it provides context around the mean.

In the context of normally distributed data, like the exercise describes, the mean is at the center of the bell curve, which makes it a helpful baseline when assessing variability with tools like standard deviation and z-score.