Lesson 17: Inference for Proportions

November 1, 2017

Review:

• t-distribution for small samples
• Exam 3 on Wed, Nov 15

Presentation:

• How to Estimate a Population Proportion
  • Conduct a Simple Random Sample (SRS) of size n
  • Record the count, x, of “successes”, e.g., # of voters supporting candidate A
  • Calculate the sample proportion, p-hat = x/n
  • If n is sufficiently large (≥30), we can assume p-hat is Normally distributed
  • Estimate of the population proportion mean, μ = p-hat
  • Estimate of the population proportion std dev, σ = √(p*(1-p)/n)
  • Estimate margin of error, m = z*σ  (use z = 1.96 for 95% confidence)
  • Estimate with 95% confidence interval is p-hat ± m
• How to do a significance test for a Proportion
  • Step 1. Setup the null hypothesis (Ho: p=0.5) and alternate hypothesis (Ha: p≠ 0.5)
  • Step 2. Calculate the test statistic, z = (p-hat – p)/√(p*(1-p)/n)
  • Step 3. Find the P-value (probability of obtaining result by chance), use Z table
  • Step 4. Interpret results, compare P-value to α = 0.05
• Video
• Example 8.1 on p. 489 – Binge Drinking Survey
  • n = 13,819
  • x = 3,140
  • p-hat = 3140/13819 = 0.227
  • standard deviation = √(p-hat*(1-p-hat)/n) = 0.00356
  • p-hat ± z*√(p-hat*(1-p-hat)/n) = 0.227 ± 1.96*(0.00356) = 0.227 ± 0.007
  • {0.22,0.234}
• Example 8.3 on p. 494 – Work Stress Survey
  • National Survey of Restaurant Workers: p = 0.75
  • Sample n = 100, x = 68, p-hat = 0.68
  • (1) Ho: p = 0.75, Ha: p ≠ 0.75
  • (2) Z = (p-hat – p)/√(p*(1-p)/n)  = (0.68 – 0.75)/√(0.75*(0.25)/100) = -1.62
  • (3) P-value = 2*P(Z≤-1.62) = 2*(0.0526) = 0.1052
  • (4) 0.1052 > 0.05  therefore  Fail to Reject Ho

Assignment:

1. A random sample of 2,454 12th-grade students were asked the following question: Taking all things together, how would you say things are these days – would you say you’re happy or not too happy? Of the responses, 2,098 students selected happy. Determine the sample proportion of students who responded they were happy and calculate a 95% confidence interval for the population proportion of 12th-grade students who are happy.
2. Currently, mothers in North America are advised to put babies to sleep on their backs. This recommendation has reduced the number of cases of sudden infant death syndrome (SIDS). However, it is a likely cause of another problem, i.e., flat spots on babies’ heads. A study of 440 babies aged 7 – 12 weeks found that 46.6% had flat spots on their heads. Calculate a 95% confidence interval for the proportion of babies in this age group that have flat spots.
3. An online article claims that 90% of American households in which a computer is owned have access to the Internet. However, an Internet provider questioned the claim. The Internet provider felt the percentage should be higher. A phone survey contacted 1,910 households in which a computer was owned and respondents were asked if they could access the Internet from their home. A total of 1,816 of the households responded yes. Conduct a hypothesis test to determine if the proportion of households who answered yes is significantly greater than 90% (0.90).