One way to check on how representative a survey is of the population from which

Question

One way to check on how representative a survey is of the population from which it was drawn is to compare various characteristics of the sample with the population characteristics. A typical variable used for this purpose is age. The 2010 GSS of the American adult population found a mean age of 49.28 and a standard deviation of 17.21 for its sample of 4,857 adults. Assume that we know from census data that the mean age of all American adults is 37.2. Use this information to answer these questions.

a. State the research and the null hypotheses for a two-tailed test.

b. Calculate the t statistic and test the null hypothesis at the .001 significance level. What did you find?

c. What is your decision about the null hypothesis? What does this tell us about how representative the sample is of the American adult population?

Explanation / Answer

Here, we use the one sample t-test. The One-Sample t-Test determines whether the sample mean is statistically different from a known or hypothesized population mean. The One-Sample t-Test is a parametric test. Since here the standard deviation is less than half of the mean, therefore, the assumption for normality is valid and we can use the parametric test.

a, Null hypothesis

The null hypothesis (H0) and (two-tailed) alternative hypothesis (H1) of the one sample T test can be expressed as:

H0: µ = x  ("the sample mean is equal to the [proposed] population mean") i.e in 2010 the GSS sample data mean age 49.28 is the representative of the American adult population obtained from Census data for the variable age 37.2

H1: µ x  ("the sample mean is not equal to the [proposed] population mean") i.e the sample data of 4857 adults for variable age is not representative of the American adults' data obtained from census

where µ is a constant proposed for the population mean and x is the sample mean.

b, calculate the t-statistics

t = x - µ/ standard error (sd/n)

standard error = 17.21/ 4857 = 17.21/ 69.692 = 0.246

t = 49.28 - 37.2 / 0.24

= 12.08/ 0.246 = 49.1

t-statistics = 49.1

To test the null hypothesis, the critical value we need to calculate

Critical value, for the at the .001 significance level, two-sided, degree of freedom (4857-1 = 4856) =   3.29

using the weblink (http://www.danielsoper.com/statcalc/calculator.aspx?id=10)

As the t-value is greater than critical value 3.29, therefore, null hypothesis is rejected

c. Since null hypothesis is rejected and observed t-statistics is very high as compared with the critical value, therefore, the age of the selected sample is much higher than the population age, therefore, may not be the representative of the US population.

STATA output is also given below

. ttesti 4857 49.28 17.21 37.2, level(99.9)

One-sample t test
------------------------------------------------------------------------------
| Obs Mean Std. Err. Std. Dev. [99.9% Conf. Interval]
---------+--------------------------------------------------------------------
x | 4857 49.28 .2469431 17.21 48.46693 50.09307
------------------------------------------------------------------------------
mean = mean(x) t = 48.9182
Ho: mean = 37.2 degrees of freedom = 4856

Ha: mean < 37.2 Ha: mean != 37.2 Ha: mean > 37.2
Pr(T < t) = 1.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 0.0000

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