Testing Claims about a Population Proportion

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Lesson 10 - 4

• Testing Claims about a Population Proportion

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Objectives

• Test a claim about a population proportion using
  the normal model
• Test a claim about a population proportion using
  the binomial probability distribution

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Vocabulary

• None new

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Requirements to test, population proportion

• Simple random sample
• n 0.05N to keep binomial vs hypergeometric
• np0(1-p0) 10 for normal approximation of
  binomial

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Steps to test population proportion

• Classical or P-value
• Test Feasible (the requirements listed before)
• Determine null and alternative hypothesis (and
  type of test two tailed, or left or right
  tailed)
• Select a level of significance a based on
  seriousness of making a Type I error
• Calculate the test statistic
• Determine the p-value or critical value using
  level of significance (hence the critical or
  reject regions)
• Compare the critical value with the test
  statistic (also known as the decision rule)
• State the conclusion

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P-Value is thearea highlighted
z0
-z0
z0
z0
Critical Region
Reject null hypothesis, if Reject null hypothesis, if Reject null hypothesis, if
P-value lt a P-value lt a P-value lt a
Left-Tailed Two-Tailed Right-Tailed
z0 lt - za z0 lt - za/2 or z0 gt za/2 z0 gt za
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Example 1 Hypothesis Test

• Nexium is a drug that can be used to reduce the
  acid produced by the body and heal damage to the
  esophagus due to acid reflux. Suppose the
  manufacturer of Nexium claims that more than 94
  of patients taking Nexium are healed within 8
  weeks. In clinical trials, 213 of 224 patients
  suffering from acid reflux disease were healed
  after 8 weeks. Test the manufacturers claim at
  the a0.01 level of significance.

H0 healed .94
n lt 0.05P assumed (P gt 5000 in US!!)
Ha healed gt .94
np(1-p) gt 10 checked224(.94)(.06) 12.63
One-sided test
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Example 1 Hypothesis Test

0.950893 0.94 Test Statistic z0
------------------------- 0.6865
?0.94(0.06)/224
a 0.01 so one-sided test yields Za 2.33
Since Z0 lt Za, we fail to reject H0 therefore
there is insufficient evidence to support
manufacturers claim
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Example 2 Binomial Probability

• According to USDA, 48.9 of males between 20 and
  39 years of age consume the minimum daily
  requirement of calcium. After an aggressive Got
  Milk campaign, the USDA conducts a survey of 35
  randomly selected males between 20 and 39 and
  find that 21 of them consume the min daily
  requirement of calcium. At the a 0.1 level of
  significance, is there evidence to conclude that
  the percentage consuming the min daily
  requirement has increased?

H0 min daily 0.489
n lt 0.05P assumed (P gt 700 in US!!)
Ha min daily gt 0.489
np(1-p) gt 10 failed 35(.489)(.511) 8.75
One-sided test
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Example 2 Binomial Probability
Since the sample size is too small to estimate
the binomial with a z-distribution, we must fall
back to the binomial distribution and calculate
the probability of getting this increase purely
by chance.
P-value P(x 21) 1 P(x lt 21) 1 P(x
20) (discrete)
1 P(x 20) is 1 binomcdf(35, 0.489, 20)
(n, p, x)
P-value 0.1261 which is greater than a, so we
fail to reject the null hypothesis (H0)
insufficient evidence to conclude that the
percentage has increased
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Using Your Calculator

• Press STAT
• Tab over to TESTS
• Select 1-PropZTest and ENTER
• Entry p0, x, and n from given data
• Highlight test type (two-sided, left, or right)
• Highlight Calculate and ENTER
• Read z-critical and p-value off screen

From first problemz0 0.686 and p-value
0.2462 Since p gt a, then we fail to reject H0
insufficient evidence to support manufacturers
claim.
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Summary and Homework

• Summary
• We can perform hypothesis tests of proportions in
  similar ways as hypothesis tests of means
• Two-tailed, left-tailed, and right-tailed tests
• The normal distribution or the binomial
  distribution should be used to compute the
  critical values for this test
• Homework
• pg 550 552 1, 2, 6, 12, 17, 26