# Tutoring statistics, rules of thumb can be key. The tutor mentions a real-life application of the 68-95-99 rule.

With hot, dry weather comes also the job of irrigation, for those who choose to do so. My wife wants a nice yard this year. Being a house-husband, my answer is “yes.”

This morning I was out observing the sprinkler’s reach, not wanting it to touch the sidewalk, if possible. Of course, it occasionally did.

Making some quick observations, I surmised the following:

Sprinkler’s mean reach: 8ft.

68% of the time it reaches about 6 inches more or less than the mean.

95% of the time if reaches about 10 inches either way from the mean.

99.7% of the time it reaches about 14 inches beyond mean, which never happened this morning, suggested the sidewalk. I estimated the sprinkler had done about 300 oscillations. So just under 1 in 300 times, the reach might be that.

I wondered: Can I model the standard deviation from those observations?

The 68-95-99 rule states that, for a normal population with mean μ and standard deviation σ,

68% of the measurements fall within μ±σ

95% within μ±1.96σ

99.7% within μ±2.75σ

Imagining a standard deviation of 6 inches, the observations don’t fit well. With μ=8, σ=5, however, they fall out

68% within 8ft±5inches (est. 6 inches)

95% within 8ft±9.8inches (est. 10 inches)

99.7% within 8ft±13.75inches (est 14 inches)

In this context, σ is difficult to estimate at a glance, since it lies decidedly in the wet patch. My estimate of 6 inches for σ could easily be much better imagined as 5 inches.

Cheers:)

Source:

Harnett, Donald and James L. Murphy. Statistical Analysis. Don Mills: Addison-Wesley, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring statistics, you might imagine everyday situations. The tutor brings up one.

Let’s imagine we have two mile runners. Runner 1, called R1, has mean time 4:45, with standard deviation 10s; R2 has mean time 5:00 with standard deviation 12s.

In any given race, give the probability R1 will beat R2.

Solution:

First, we convert the mile times to seconds: R1’s mean is 285s, while R2’s is 300.

The expected difference between R2 and R1’s time is 300-285=15.

We can’t add standard deviations, but rather variances: 10^2 + 12^2 = 244. The standard deviation of the difference is then 244^0.5 = 15.6.

The standardized statistic is z = (x-15)/15.6. We wonder p(x>0), which means p(z>-0.96). From the z-table, the answer is 0.8315.

So, R1 should beat R2 about 83% of the time.

Source:

Harnett, Donald L. and James L. Murphy. Statistical Analysis for Business and Economics. Don Mills: Addison-Wesley, 1993.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring stats, you deal with p-values. The tutor shows an Excel connection.

In my posts here and here I mention p-values.

Example: Using Excel, get a two-tailed p-value for z=2.4

Solution: Using symmetry, it’s best to get the cumulative z-probability for -2.4, then double it:

=2*normsdist(-2.4)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring statistics, distributions are of constant interest. The tutor brings up ten points about the chi-square distribution.

The chi-square distribution may not be discussed much in a first-level stats course. It’s used to estimate or evaluate variance, rather than central tendency. Here are ten facts about the chi-square distribution:

1. Typically, Χ2v is used to denote a chi-square variable or distribution with v degrees of freedom.
2. Its parameter used for finding its values in tables is the degrees of freedom, typically referred to as v or just n-1, where n is the number of values in the sample.
3. Its expected value is v.
4. Its variance is 2v.
5. It’s not symmetrical, but skewed right.
6. Since a chi-square random variable is calculated from summing squares, it can’t be negative.
7. (n-1)s22 has a chi-square distribution with n-1 degrees of freedom. In this context, s2 is the sample variance, while σ2 is the true population variance.
8. Σ(observed-predicted)2/predicted, for predicted values using a model, follows a chi-square distribution.
9. The chi-square distribution is used to estimate or test population variance.
10. The chi-square distribution is used to test goodness-of-fit between a model and a sample.

I’ll be talking more about the uses of the chi-square distribution.

Source:

Harnett, Donald L. and James L. Murphy. Statistical Analysis for Business and Economics, third edition. Don Mills: Addison-Wesley, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring statistics, linear regression is perennial. The tutor mentions an assumption it includes.

When appropriate, linear regression models data by the equation

y = a + bx + e,

e being an error term due to variability.

An inherent assumption of linear regression modelling is that the error term, e, does not depend on the actual data value, x.

In many lab environments, the assumption that error magnitude does not depend on the measurement’s magnitude makes sense. For instance, measuring with a ruler, the error is often set to ± 0.5mm, regardless of the length measured.

For some types of data, however, the measurement’s magnitude seems to impact its error magnitude. An example might be inventory counting. One imagines that, counting only three items, the error would likely be 0. Counting a thousand, however, would more likely yield an observation a few off from the real number present, and so on.

Perhaps the point is that the data has to be measured or observed, which itself brings error, perhaps dependent on the size of the measurement itself.

Source:

Harnett, Donald L. and James L. Murphy. Statistical Analysis for Business and Economics. Don Mills: Addison-Wesley, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring statistics, mean is used more than median. The tutor points out a reason why.

In my post from November 30, 2014, I write about why the median might be preferred to the mean in some cases. For instance, the median tends to be less sensitive to outliers. Why, then, is the mean typically preferred in academic statistics?

The estimator of the median, it turns out, has higher variance than that of the mean. In particular,

var (sample median) = (π/2)*var(x)

Therefore, if you want to estimate the true expected value of the population, a sample size of 157 is needed to estimate it as precisely from the median as you can from a sample size of 100 using the mean.

HTH:)

Source:

Harnett, Donald L. and James L. Murphy. Statistical Analysis for Business and Economics. Don Mills: Addison Wesley, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring statistics, rules of usage are key. The tutor shares one about when the normal approximation to the binomial distribution can be used.

The binomial distribution imagines a series of n trials, each with probability p of success and q=(1-p) of failure. The number of successes in n trials is the random variable x.

Apparently, the binomial random variable x can be approximated to normal if np≥5 and also nq≥5.

HTH:)

Source:

PennState Eberly College of Science

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring statistics, the concept of skew arises. The tutor gives some brief ideas about skew.

The normal distribution isn’t skewed, because its left and right tails are simply mirror images of each other.

When a distribution’s left tail is longer than its right, we say that distribution is negatively skewed. In such a case you get a long left tail that ramps up to a clump of values on the right.

The opposite is true for a positively skewed distribution; most of the population is clumped to the left, but a long tail extends to the right.

Therefore, skew describes the location of the distribution’s longer tail: negative skew means long tail to the left (the negative) side, whereas positive skew means long tail to the right (positive) side.

HTH:)

Source:

Harnett, Donald L. and James L. Murphy. Statistical Analysis for Business and Economics, 3rd ed. Don Mills: Addison-Wesley, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring statistics, regression is an important topic. The tutor defines multiple linear regression.

Simple linear regression models the dependent variable by a linear connection to the independent:

y = a + bx

Multiple linear regression models the dependent variable by a linear connection across several independent variables:

y = a + b1x1 + b2x2 + b3x3 + ….

Source:

Harnett, Donald L. and James L. Murphy. Statistical Analysis for Business and Economics. Don Mills: Addison-Wesley, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring math, you see all kinds of scientific calculators. The tutor continues to praise the Casio fx-991ES PLUS C.

If you were to be marooned on an island with a single calculator…which one would you prefer? The question is absurd, of course. Yet, I believe I know my answer: the Casio fx-991ES PLUS C. It has seemingly endless functionality for a little scientific calculator.

Today I noticed that it calculates binomial pdf.

Example: Using the Casio fx-991ES PLUS C, find the probability of 20 successes in 35 trials when p=0.44.

Solution:

1. Press MODE, then arrow down
2. Select 3
3. Select 4
4. For a single query, select 2
5. It will ask for x (the number of successes). Key in 20 =
6. Next, it will ask for N (the number of trials). Key in 35 =
7. Now it will ask for p (the probability of success each time). Key in .44 =