Lab 9: Inference for binomial data

With solutions

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This lab covers inference for population proportions from binomial (two-category) data. Examples use the NHANES survey data.

load('data/nhanes500.RData')

Point estimation

How common is diabetes among U.S. adults?

For categorical data, point estimates of population proportions are computed by tabulating counts and normalizing by the sample size:

# tabulate counts
diabetes.tbl <- table(nhanes$diabetes)
diabetes.tbl

Yes  No 
 57 443 
# compute sample proportions
diabetes.tbl |> prop.table()

  Yes    No 
0.114 0.886

Diabetes prevalence among U.S. adults is estimated to be 11.4%.

How common is sleep trouble among U.S. adults?

CautionYour turn 1 [L9]

The sleeptrouble variable records whether the participant experiences sleep trouble. Tabulate the counts and store the table as sleep.tbl. Then compute sample proportions and estimate the prevalence of sleep trouble.

# tabulate and store
sleep.tbl <- table(nhanes$sleeptrouble)
sleep.tbl

Yes  No 
139 361 
# compute sample proportions
sleep.tbl |> prop.table()

  Yes    No 
0.278 0.722

An estimated 27.8% of U.S. adults experience sleep trouble.

Testing a binomial proportion

Is diabetes prevalence really 10%?

To test whether a population proportion equals a hypothesized value, use prop.test(...). Pass a table of counts and specify the hypothesized proportion p:

# test of binomial proportion (table input)
diabetes.tbl |> prop.test(p = 0.1)

    1-sample proportions test with continuity correction

data:  diabetes.tbl, null probability 0.1
X-squared = 0.93889, df = 1, p-value = 0.3326
alternative hypothesis: true p is not equal to 0.1
95 percent confidence interval:
 0.08814952 0.14594579
sample estimates:
    p 
0.114

The output includes the test statistic, p-value, and a confidence interval.

The data do not provide evidence that diabetes prevalence among U.S. adults differs from 10% (\(\chi^2\) = 0.94 on 1 degree of freedom, p = 0.3326). With 95% confidence, diabetes prevalence is estimated to be between 8.81% and 14.59%.

Is one in five U.S. adults affected by sleep trouble?

CautionYour turn 2 [L9]

Test whether one in five U.S. adults experience sleep trouble. Store the full test result as sleep.rslt. Report the test result and confidence interval in context following conventional style.

# test of binomial proportion
sleep.rslt <- sleep.tbl |> prop.test(p = 0.2)
sleep.rslt

    1-sample proportions test with continuity correction

data:  sleep.tbl, null probability 0.2
X-squared = 18.528, df = 1, p-value = 1.674e-05
alternative hypothesis: true p is not equal to 0.2
95 percent confidence interval:
 0.2395872 0.3198838
sample estimates:
    p 
0.278

The data provide evidence that the proportion of U.S. adults who experience sleep trouble differs from 0.2 (\(\chi^2\) = 18.53 on 1 degree of freedom, p < 0.0001). With 95% confidence, between 23.96% and 31.99% of U.S. adults experience sleep trouble.

Alternatively, prop.test(...) can accept the count and sample size directly instead of a table. For the diabetes example, 57 out of 500 participants are diabetic:

# test of binomial proportion (count input)
prop.test(x = 57, n = 500, p = 0.1)

    1-sample proportions test with continuity correction

data:  57 out of 500, null probability 0.1
X-squared = 0.93889, df = 1, p-value = 0.3326
alternative hypothesis: true p is not equal to 0.1
95 percent confidence interval:
 0.08814952 0.14594579
sample estimates:
    p 
0.114

The results are identical.

Does more than half the population experience no sleep trouble?

CautionYour turn 3 [L9]

Using the count and sample size directly (i.e., prop.test(x = ..., n = ..., p = ...)), test whether 80% of U.S. adults are free of sleep trouble. Store the result as nosleep.rslt. Report the result in context.

# test using count input
nosleep.rslt <- prop.test(x = 361, n = 500, p = 0.8)
nosleep.rslt

    1-sample proportions test with continuity correction

data:  361 out of 500, null probability 0.8
X-squared = 18.528, df = 1, p-value = 1.674e-05
alternative hypothesis: true p is not equal to 0.8
95 percent confidence interval:
 0.6801162 0.7604128
sample estimates:
    p 
0.722

The data provide evidence that the proportion of U.S. adults free of sleep trouble differs from 80% (\(\chi^2\) = 18.53 on 1 degree of freedom, p < 0.0001). With 95% confidence, between 68.01% and 76.04% of U.S. adults are free of sleep trouble.

Adjusting the confidence level

The default confidence level is 95%. To change it, add a conf.level argument:

# 99% confidence interval
diabetes.tbl |> prop.test(p = 0.1, conf.level = 0.99)

    1-sample proportions test with continuity correction

data:  diabetes.tbl, null probability 0.1
X-squared = 0.93889, df = 1, p-value = 0.3326
alternative hypothesis: true p is not equal to 0.1
99 percent confidence interval:
 0.0814852 0.1568967
sample estimates:
    p 
0.114

With 99% confidence, diabetes prevalence among U.S. adults is estimated to be between 8.10% and 15.66%.

How precise is the estimate of sleep trouble prevalence at the 90% level?

CautionYour turn 4 [L9]

Construct a 90% confidence interval for the proportion of U.S. adults who experience sleep trouble by testing against the same hypothesis as Your turn 2 but with a different confidence level. Store the result as sleep90.rslt. Interpret the interval in context.

# 90% confidence interval
sleep90.rslt <- sleep.tbl |> prop.test(p = 0.2, conf.level = 0.90)
sleep90.rslt

    1-sample proportions test with continuity correction

data:  sleep.tbl, null probability 0.2
X-squared = 18.528, df = 1, p-value = 1.674e-05
alternative hypothesis: true p is not equal to 0.2
90 percent confidence interval:
 0.2453476 0.3131143
sample estimates:
    p 
0.278

With 90% confidence, between 24.60% and 31.27% of U.S. adults experience sleep trouble.

Directional tests

A directional (one-sided) test is performed by specifying alternative = 'greater' or alternative = 'less'. For example, to test whether diabetes prevalence exceeds 10%:

# upper-sided test
diabetes.tbl |> prop.test(p = 0.1, alternative = 'greater')

    1-sample proportions test with continuity correction

data:  diabetes.tbl, null probability 0.1
X-squared = 0.93889, df = 1, p-value = 0.1663
alternative hypothesis: true p is greater than 0.1
95 percent confidence interval:
 0.09176378 1.00000000
sample estimates:
    p 
0.114

The data do not provide evidence that diabetes prevalence among U.S. adults exceeds 10% (\(\chi^2\) = 0.94 on 1 degree of freedom, p = 0.1663). With 95% confidence, diabetes prevalence is estimated to be at least 9.18%.

Does more than a quarter of U.S. adults experience sleep trouble?

CautionYour turn 5 [L9]

Test whether sleep trouble prevalence exceeds 25%. Store the result as sleep.upper.rslt. Report the test result and lower confidence bound in context.

# upper-sided test
sleep.upper.rslt <- sleep.tbl |>
  prop.test(p = 0.25, alternative = 'greater')
sleep.upper.rslt

    1-sample proportions test with continuity correction

data:  sleep.tbl, null probability 0.25
X-squared = 1.944, df = 1, p-value = 0.08162
alternative hypothesis: true p is greater than 0.25
95 percent confidence interval:
 0.2453476 1.0000000
sample estimates:
    p 
0.278

The data do not provide evidence that sleep trouble prevalence exceeds 25% (\(\chi^2\) = 1.94 on 1 degree of freedom, p = 0.0816). With 95% confidence, sleep trouble prevalence is estimated to be at least 24.53%.

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