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Pediatric asthma survey

  1. Pediatric asthma survey, n = 50. Suppose that asthma affects 1 in 20 children in a population. You take an SRS of 50 children from this population. Can the normal approximation to the binomial be applied under these conditions? If not, what probability model can be used to describe the sampling variability of the number of asthmatics?

  2. Misconceived hypotheses. What is wrong with each of the following hypothesis statements?
    a) H0: μ = 100 vs. Ha: μ ≠ 110
    b) H0: x̄ = 100 vs. Ha: x̄ < 100 or could write as H0: x̄ >= 100 vs. Ha: x̄ < 100
    c) H0: p^ = 0.50 vs. Ha: p^ ≠ 0.50

  3. Patient satisfaction. Scores derived from a patient satisfaction survey are Normally distributed with μ = 50 and σ = 7.5, with high scores indicating high satisfaction. An SRS of n = 36 is taken from this population.
    a) What is the standard error (SE) of x for these data?
    b) We seek to discover if a particular group of patients comes from this population in which μ = 50. Sketch the curve that describes the sampling distribution of the sample mean under the null hypothesis. Mark the horizontal axis with values that are ±1, ±2, and ±3 standard errors above and below the mean.
    c) Suppose in a sample of n = 36 from this particular group of patients the mean value of x is 48.8. Mark this finding on the horizontal axis of your sketch. Then compute a z statistic for this scenario and make sure it matches your sketch.
    d) What is the two-sided alternative hypothesis for this scenario?
    e) Find the corresponding p-value for your z-statistic using Table B.
    f) Draw a conclusion for this study scenario based on your results.

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Sample Answer

 

 

 

1. Normal Approximation to the Binomial:

No, the normal approximation to the binomial cannot be applied under these conditions.

Here’s why:

  • Rule of Thumb: The normal approximation is generally considered valid when both np and nq are greater than or equal to 5.
  • In this case:
    • n (sample size) = 50
    • p (probability of having asthma) = 1/20 = 0.05
    • np = 50 * 0.05 = 2.5
    • nq = 50 * (1 – 0.05) = 47.5

Since np (2.5) is less than 5, the normal approximation is not appropriate.

Full Answer Section

 

 

 

Probability Model for Sampling Variability:

The appropriate model for this scenario is the Binomial Distribution. The binomial distribution describes the probability of getting a certain number of successes (asthmatics) in a fixed number of trials (sample size) when each trial has two possible outcomes (asthma or no asthma) with a constant probability (1/20).

  1. Misconceived Hypotheses:
  2. a) H0: μ = 100 vs. Ha: μ ≠ 110
  • Problem:The null hypothesis (H0) specifies an exact value for the population mean (μ). In hypothesis testing, H0 usually represents the status quo or no effect.

**b) H0: x̄ = 100 vs. Ha: x̄ < 100 or could write as H0: x̄ >= 100 vs. Ha: x̄ < 100  **

  • Problems:
    • Both H0 and Ha involve the sample mean (x̄), which is unknown for the population.
    • The alternative hypothesis (Ha) is directional (only less than) and doesn’t consider the possibility of a difference in either direction.
  1. c) H0: p^ = 0.50 vs. Ha: p^ ≠ 0.50
  • No problem:This is a correctly written hypothesis for a proportion test. H0 states the null hypothesis that the population proportion (p^) is 0.50, while Ha is the two-sided alternative that it’s not equal to 0.50.
  1. Patient Satisfaction:
  2. a) Standard Error (SE) of the Mean:

SE(x̄) = σ / √n SE(x̄) = 7.5 / √36 = 1.25

  1. b) Sampling Distribution of the Mean:
  • Sketch a normal curve with a mean (μ) of 50 on the x-axis.
  • Mark ±1 SE (1.25), ±2 SE (2.5), and ±3 SE (3.75) on the x-axis above and below the mean.
  1. c) Sample Mean and z-statistic:
  • Mark 48.8 on the x-axis (since the sample mean is lower than the population mean).
  • Calculate the z-statistic:

z = (x̄ – μ) / SE(x̄) z = (48.8 – 50) / 1.25 = -0.96

This negative z-score corresponds to the tail region to the left of the mean on your sketch.

  1. d) Two-sided Alternative Hypothesis:

Ha: μ ≠ 50 (This is because we are interested in any difference from the population mean, not just a decrease).

  1. e) p-value:

Look up 0.96 (absolute value) on the z-table. The two-tailed p-value is approximately 0.3413.

  1. f) Conclusion:

Since the p-value (0.3413) is greater than a typical significance level (e.g., 0.05), we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean satisfaction score for this particular group of patients is statistically different from the population mean of 50.

 

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