1. Pediatric Asthma Survey

Given that asthma affects 1 in 20 children in the population, the probability of a child having asthma is ( p = \frac{1}{20} = 0.05 ). For a sample size of 50 children, the mean and variance of the binomial distribution can be calculated as ( \mu = np = 50 \times 0.05 = 2.5 ) and ( \sigma^2 = np(1-p) = 50 \times 0.05 \times 0.95 = 2.375 ). The standard deviation is ( \sigma = \sqrt{2.375} \approx 1.54 ).

Since the number of asthmatic children in the sample (2.5) is less than 10, the normal approximation to the binomial cannot be applied due to the sample being small and the success rate being low. Instead, the sampling variability of the number of asthmatics can be described using a Poisson distribution since the conditions for using the Poisson model (rare events and independent trials) are met.

2. Misconceived Hypotheses

a) The alternative hypothesis should be specific, stating whether the population mean is greater or less than the hypothesized value. It should be ( H_a: \mu \neq 100 ) or ( H_a: \mu > 100 ) depending on the research question.

b) The null hypothesis should be based on population parameters, not sample statistics. It should be formulated as ( H_0: \mu = 100 ) against an alternative hypothesis that specifies a direction, such as ( H_a: \mu < 100 ).

c) The null hypothesis should be based on a specific value and not estimated from the data. It should be ( H_0: p = 0.50 ) against an alternative hypothesis that specifies a deviation from this value, such as ( H_a: p \neq 0.50 ).

3. Patient Satisfaction

a) The standard error (SE) of the sample mean can be calculated as ( SE = \frac{\sigma}{\sqrt{n}} = \frac{7.5}{\sqrt{36}} = 1.25 ).

b) Under the null hypothesis where μ = 50, the sampling distribution of the sample mean follows a normal distribution with mean μ = 50 and standard error SE = 1.25. The curve would be bell-shaped with values marked at ±1.25, ±2.5, and ±3.75 around the mean.

c) Marking x̄ = 48.8 on the horizontal axis, the z-statistic can be calculated as ( z = \frac{48.8 – 50}{1.25} = -1.44 ). This value should align with the sketch.

d) The two-sided alternative hypothesis for this scenario would be ( H_a: μ \neq 50 ).

e) Finding the corresponding p-value for a z-statistic of -1.44 from Table B will provide information on the significance level of the sample mean being 48.8.

f) Based on the p-value obtained, a conclusion can be drawn regarding whether the particular group of patients significantly differs from the general population in terms of satisfaction scores at a confidence level chosen by comparing the p-value to the significance level α.