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This module explains how to construct a confidence interval estimate for an unknown population mean when the population standard deviation is unknown, using the Student-t distribution. This module has been revised from the original module m16959 by S. Dean and Dr. B. Illowsky in the textbook collection Collaborative Statistics to include step by step solutions for all examples.

In practice, we rarely know the population standard deviation . In the past, when the sample size was large, this did not present a problem to statisticians. They used thesample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in theconfidence interval.

William S. Gossett (1876-1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very fewsamples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distributionfor the calculation; he found that the actual distribution depends on the sample size. This problem led him to "discover" what is called the Student-t distribution . The name comes from the fact that Gosset wrote under the pen name "Student."

Up until the mid 1990s, statisticians used the normal distribution approximation for large sample sizes and only used the Student-t distribution for sample sizes of at most 30.With the common use of graphing calculators and computers, the practice is to use the Student-t distribution whenever s is used as an estimate for σ .

If you draw a simple random sample of size n from a population that has approximately a normal distribution with mean μ and unknown population standard deviation σ and calculate the t-score t = x - μ ( s n ) , then the t-scores follow a Student-t distribution with n - 1 degrees of freedom . The t-score has the same interpretation as the z-score    . It measures how far x is from its mean μ . For each sample size n , there is a different Student-t distribution.

The degrees of freedom , n - 1 , come from the calculation of the sample standard deviation s . In Chapter 2, we used n deviations ( x - x values ) to calculate s . Because the sum of the deviations is 0, we can find the last deviation once we know theother n - 1 deviations. The other n - 1 deviations can change or vary freely. We call the number n - 1 the degrees of freedom (df).

    Properties of the student-t distribution

  • The graph for the Student-t distribution is similar to the Standard Normal curve.
  • The mean for the Student-t distribution is 0 and the distribution is symmetric about 0.
  • The Student-t distribution has more probability in its tails than the Standard Normal distribution because the spread of the t distribution is greater than the spread of the Standard Normal. So the graph of the Student-t distribution will be thicker in the tails and shorter in the center than the graph of the Standard Normal distribution.
  • The exact shape of the Student-t distribution depends on the "degrees of freedom". As the degrees of freedom increases, the graph Student-t distribution becomes more like the graph of the Standard Normal distribution.
  • The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ . In the real world, however, as long as the underlying population is large and bell-shaped, and the data are a simple random sample, practitioners often consider the assumptions met.

Calculators and computers can easily calculate any Student-t probabilities. The TI-83,83+,84+ have a tcdf function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.

For the TI-84+ we will use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm.The invT command requires two inputs: invT(area to the left, degrees of freedom) The output is the t-score that corresponds to the area we specified.

The TI-83 and TI-83+ do not have the invT command but you can download an invT program from your instructor that is easy to use. (The TI-89 has an inverse T command.)

A probability table for the Student-t distribution can also be used. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student-t distribution.) When using t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails.

    The notation for the student-t distribution is (using t as the random variable) is

  • T ~ t df where df = n - 1 .
  • For example, if we have a sample of size n=20 items, then we calculate the degrees of freedom as df=n−1=20−1=19 and we write the distribution as T ~ t 19

If the population standard deviation is not known , the error bound for a population mean is:

  • EBM = t α 2 ( s n )
  • t α 2 is the t-score with area to the right equal to α 2
  • use df = n - 1 degrees of freedom
  • s = sample standard deviation

Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects withthe results given below. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) fromwhich you took the data.

  • 8.6
  • 9.4
  • 7.9
  • 6.8
  • 8.3
  • 7.3
  • 9.2
  • 9.6
  • 8.7
  • 11.4
  • 10.3
  • 5.4
  • 8.1
  • 5.5
  • 6.9
  • The solution is shown step by step.

To find the confidence interval, you need the sample mean, x , and the EBM.

x = 8.2267 s = 1.6722 n = 15

df = 15 - 1 = 14

CL = 0.95 so α = 1 - CL = 1 - 0.95 = 0.05

α 2 = 0.025 t α 2 = t .025

The area to the right of t .025 is 0.025 and the area to the left of t .025 is 1−0.025=0.975

t α 2 = t .025 = 2.14 using invT(.975,14) on the TI-84+ calculator.

EBM = t α 2 ( s n )

EBM = 2.14 ( 1.6722 15 ) = 0.924

x - EBM = 8.2267 - 0.9240 = 7.3

x + EBM = 8.2267 + 0.9240 = 9.15

The 95% confidence interval is (7.30, 9.15) .

We estimate with 95% confidence that the true population average sensory rate is between 7.30 and 9.15.

Note: When calculating the error bound, a probability table for the Student-t distribution can also be used to find the value of t. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row); the t-score is found where the row and column intersect in the table.

Questions & Answers

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Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
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To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
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8m/s²
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Source:  OpenStax, Collaborative statistics: custom version modified by r. bloom. OpenStax CNX. Nov 15, 2010 Download for free at http://legacy.cnx.org/content/col10617/1.4
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