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?
P=1/20= 0.05
N=50
Normal approximation to binomial is valid when np >5 and nq > 5
Here, np=2.5 and nq=47.5 hence Normal approximation to binomial is NOT valid.
Binomial distribution is applicable here, X follows B(n=50, p=0.05). X is defined as asthma affects
2. Fill in the blanks. A particular random sample of n observations can be used to calculate a sample proportion. The count of successes in the sample will vary according to a (a) _______discrete_____________ probability distribution with parameters n and (b) _______p_____________. When the sample is large, the number of success will vary according to a normal distribution with μ = (c) _______np_____________ and σ = (d) _________sqrt(npq)___________. At the same time, the sampling distribution of proportion in large samples will be Normally distributed with mean p with standard deviation equal to (e) __________sqrt(pq/n)__________.
3. Misconceived hypotheses. What is wrong with each of the following hypothesis statements?
a) H0: μ = 100 vs. Ha: μ ≠ 110
The null hypothesis is set as per the claim. Here the following possibilities which can be true:
H0: μ = 100 vs. Ha: μ ≠ 110
Or H0: μ = 110 vs. Ha: μ ≠ 110
b) H0: = 100 vs. Ha: < 100
The population parameter (mean) to be analyzed is missing. Here the following possibilities which can be true:
H0: μ = 100 vs. Ha: μ < 100
c) H0: = 0.50 vs. Ha: ≠ 0.50
The population parameter (proportion) to be analyzed is missing. Here the following possibilities which can be true:
H0: p = 0.5 vs. Ha: p ≠ 0.50
4. 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 SE of x̅ for these data?
SE = sd/sqrt(n) = 7.5/sqrt(36) = 1.25
b) We seek evidence against the hypothesis that a particular group of patients comes from a population in which μ = 50. Sketch the curve that describes the sampling distribution of x̅ under the null hypothesis. Mark the horizontal axis with values that are ±1, ±2, and ±3 standard errors above and below the mean.Data set 1: patient’s satisfaction (b)
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c) Suppose a sample of n = 36 finds an x̅ of 48.8. Mark this finding on the horizontal axis of your sketch. Then calculate the zstat for the result. Does this observation provide strong evidence against H0?Result 2: Normal distribution table C
Ho: u = 50
H1: u =/= 50
Z= (mean-u)/SE = (48.8-50)/1.25 = -0.96
Z(a/2) = z(0.05/2) = 1.96
Since, |z| < z(a/2), I fail to reject ho at 5% level of significance and conclude that u = 50
5. Patient satisfaction (sample mean of 48.8). In Exercise 4, you tested H0: μ = 50 based on a sample of n = 36 showing = 48.8. The population had standard deviation σ = 7.5.
a) What is the one-sided alternative hypothesis for this test?
H1: u > 50
b) Calculate the z-statistic for the test.
Z= (mean-u)/SE = (48.8-50)/1.25 = -0.96
c) Convert the zstat to a P-value, and interpret the results.
P(z>-.96)
=1-P(z<=-.96)
=1-NORMSDIST(-0.96)
=0.83
Since, p-value is greater than alpha (0.05), I fail to reject ho at 5% level of significance and conclude that u = 50
