7.2: A Single Population Mean using the Student t Distribution

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A Single Population Mean using the Student t Distribution

jkesler

[latexpage]

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 the sample 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 the confidence interval.

William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. 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 distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to “discover” what is called the Student’s t-distribution. The name comes from the fact that Gosset wrote under the pen name “Student.”

Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and used the Student’s t-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student’s 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 an approximately normal distribution with mean μ and unknown population standard deviation σ and calculate the t-score $$t=\frac{\bar x – \mu}{s/\sqrt{n}}$$ then the t-scores follow a Student’s t-distribution with n – 1 degrees of freedom. The t-score has the same interpretation as the z-score. It measures how far $ \bar x$ is from its mean μ. For each sample size n, there is a different Student’s t-distribution.

The degrees of freedom, n – 1, come from the calculation of the sample standard deviation s. In Figure G1, we used n deviations $(x – \bar{x} \text{ values})$ to calculate s. Because the sum of the deviations is zero, we can find the last deviation once we know the other 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’s t-Distribution

• The graph for the Student’s t-distribution is similar to the standard normal curve.
• The mean for the Student’s t-distribution is zero and the distribution is symmetric about zero.
• The Student’s 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’s 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’s t-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student’s 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. The size of the underlying population is generally not relevant unless it is very small. If it is bell shaped (normal) then the assumption is met and doesn’t need discussion. Random sampling is assumed, but that is a completely separate assumption from normality.

Google Sheets

Calculators and computers can easily calculate any Student’s t-probabilities. Google Sheets has the T.DIST function (along with T.DIST.2T and T.DIST.RT explained in a later chapter ) to find the probability for given values of t. The grammar for the T.DIST command is T.DIST(x, degrees of freedom, cumulative?). However for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.

For Google Sheets, you can use the T.INV function. The T.INV function works similarly to the NORM.INV function. The T.INV function requires two inputs: T.INV(area to the left, degrees of freedom) The output is the t-score that corresponds to the area we specified.

 

A probability table for the Student’s t-distribution can also be used. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row).

A Student’s t table gives t-scores given the degrees of freedom and the right-tailed probability. The table is very limited. Spreadsheets can easily calculate any Student’s t-probabilities.

The notation for the Student’s t-distribution (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 ~ t₁₉.

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

• $E = t_{\alpha/2}\cdot\frac{s}{\sqrt{n}}$
• $t_{\alpha/2}$ is the $t$-score with an area to the right equal to $\frac{\alpha}{2}$
  Note that the T.INV function in a spreadsheet expects to be given a probability to the left. Since $t_{\alpha/2}$ is the $t$-score with an area to the right, we simply give the T.INV function the value $1-\alpha/2$.
• use df = n – 1 degrees of freedom, and
• s = sample standard deviation.

The format for the confidence interval is:
$(\bar x – E, \bar x + E)$

Example 7.8

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

Solution 7.8

Google Sheets

If you copy the data from the problem into a spreadsheet, including the semi-colons, all data will be placed into a single cell which makes calculations (like the mean, standard deviation, etc) on the data difficult. Instead, use the Split Text to Columns functionality so each data point will be in a separate cell. Then use AVERAGE and STDEV.S functions to calculate the mean and standard deviation of the data.

To find the confidence interval, you need the sample mean, $\bar x$, and the margin of error, E . First, use a spreadsheet to calculate the mean, standard deviation, and note the $n=15$.

$\bar x = 8.2267, ~~s=1.6722, ~~n=15$

df = 15 – 1 = 14 CL so α = 1 – CL = 1 – 0.95 = 0.05

$\alpha=0.05$, so $t_{\alpha/2} = t_{0.025}$

The area to the right of t_0.025 is 0.025

$t_{\alpha/2} = t_{0.025}=2.14$ using T.INV(0.975,14) on a spreadsheet.

$E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}=2.14\cdot \frac{1.6722}{\sqrt{15}}$

$\bar x – E = 8.267-0.924 = 7.3$

$\bar x +E 8.267+0.924=9.15$

The 95% confidence interval is (7.3, 9.15).

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

Note

When calculating the margin of error, a probability table for the Student’s 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.

Try It 7.8

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data.

8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5

Example 7.9

The Human Toxome Project (HTP) is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. In October 2008, the scientists at HTP tested cord blood samples for 20 newborn infants in the United States. The cord blood of the “In utero/newborn” group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous system toxicity, immune system toxicity, and reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. Table 7.3 below shows how many of the targeted chemicals were found in each infant’s cord blood.

79	145	147	160	116	100	159	151	156	126
137	83	156	94	121	144	123	114	139	99

Table 7.3

Use this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an in infant’s blood.

Solution 7.9

From the sample, you can calculate $\bar x = 127.45$ and $s=25.965$. There are 20 infants in the sample, so n = 20, and df = 20 – 1 = 19.

You are asked to calculate a 90% confidence interval: CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10
$\frac{\alpha}{2} = 0.05$ so $t_{\alpha/2}=t_{0.05}$

By definition, the area to the right of t_0.05 is 0.05 and so the area to the left of t_0.05 is 1 – 0.05 = 0.95.

Use a table, calculator, or computer to find that t_0.05 = 1.729. In Google Sheets, use T.INV(0.95, 19).

$E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 1.729 \cdot \frac{25.965}{\sqrt{20}} = 10.038$

$\bar x – E = 127.45-10.038 = 117.412$

$\bar x + E = 127.45+10.038 = 137.488$

We estimate with 90% confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488.

Try It 7.9

A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in Table 7.4 below. Use this sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week.

0	3	1	20	9
5	10	1	10	4
14	2	4	4	5

Table 7.4