Confidence Interval for Mean
of a Normally Distributed Population

 TCONFINT.83p TCONFINT.86p tconfint.89p

When studying a population that is assumed to be normally distributed, such as IQ scores or heights, we can estimate the mean µ of the population by finding the sample mean Xbar of a random sample of n measurements. But since Xbar is only an estimate, we must say that µ = Xbar +/- e, where e is some appropriate margin of error.

The margin of error depends on a desired level of confidence r, such as 0.90, 0.95, 0.98, or 0.99. The range of the estimate for µ, given by [Xbar- e, Xbar+ e] is called the confidence interval.

The level of confidence r and the sample size n determine a t-score t as follows: If t(n - 1) denotes the t-distribution with n - 1 degress of freedom, then t is the value such that P(-t <= t(n - 1) <= t) = r.

The margin of error is then determined by the sample deviation Sx, the t-score t, and the sample size n and is given by e = t*Sx /Sqrt(n); thus,

µ ~ Xbar +/- t*Sx /Sqrt(n)

When the population is known to be normally distributed, then we do not need large sample sizes to use this analysis.

Using the TCONFINT Program

To execute the TCONFINT program, enter the sample size, the values of Xbar and Sx, and the desired level of confidence (in decimal). The program displays the sample mean, the margin of error, and the confidence interval.

Using a Data Set: If we have a collection of meaurements, then we can enter the data into the STAT Edit screen (LIST Edit on the 86, APPS 6 on the TI-89) in order to compute the sample mean Xbar and sample deviation Sx. We can then use the VARS screen to enter these values into the program.

Example. From a population of normally distributed IQ scores, a random sample of 25 scores yields Xbar = 98.4 and Sx = 14.2. Find a 98% confidence interval for the true mean score µ of this population.

Solution. After calling up the TCONFINT program, enter 25 for SAMPLE SIZE, 98.4 for XBAR, 14.2 for SAMPLE DEV., and .98 for CONF. LEVEL. We see that µ ~ 98.4 +/- 7.0777, that gives us a 98% confidence interval of [91.3223, 105.4777].

Exercises

1. Standardized test scores are often found to be normally distributed. The data set below is a random sample of combined SAT scores from a group of sophomores picked from a random university. Find a 95% confidence interval for the mean SAT score of sophomores at this university.

Random Collection of SAT Scores

 1270 1070 1440 880 970 1400 860 980 1220 1110 850 1080 1210 720 920 800 1160 1220 1280 1200 1030 1170 1350 1070 950 1150 1020 1120 1020 1070 910 1320

2. A random sample of size 400 from a normally distributed population yields Xbar = 980.5 and Sx = 56.3. Comapre the results of finding a 99% confidence interval for µ when using z-scores rather than t-scores.

Solutions

1. First, enter the data into the STAT Edit screen (LIST EDIT on the TI-86, APPS 6 on the TI-89), and use the 1-Var Stats command (OneVar on the TI-86 and 89) to compute the statistics. We see that n = 32, Xbar = 1088.125 and Sx ~ 177.9532.

Next, call up the TCONFINT program and enter the statistics when prompted. We find that the average SAT score is 1088.125 +/- 64.159, that gives us a 95% confidence interval of [1023.966, 1152.284].

2. Because we have a relatively large sample and the t-distributions converge in probability to the standard normal distribution, we can find a confidence interval with the ZCONFINT program that uses z-scores. Entering the data into this program, we find that µ ~ 980.5 +/- 7.251 giving an interval of (973.249, 987.751).

Using t-scores with the TCONFINT program, we obtain a 99% confidence interval of 980.5 +/- 7.258, or (973.2142, 987.7858).