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3. Suppose the annual incidence of asthma in the general population among childr...

Sep 25, 2024

3. Suppose the annual incidence of asthma in the general population among children 0- 4 years of age is 1.4% for boys and 1% for girls. (a) If 10 cases are observed over 1 year among 500 boys 0–4 years of age with smoking mothers, then test whether there is a significant difference in asthma incidence between this group and the general population using the critical-value method with a two-sided test. (b) Report p-value corresponding to your answer in (a) (c) Suppose that four cases are observed over 1 year among 300 girls 0–4 years of age with smoking mothers. Answer Problem 7.15 based on these data. (d) Report p-value corresponding to your answer in (c)

Solution by Steps

step 1

To test whether there is a significant difference in asthma incidence between the observed group and the general population, we first calculate the expected number of cases among 500 boys. The incidence for boys is 1.4%, so the expected cases are:

E = 500 \times 0.014 = 7

step 2

We will use a two-sided test for the hypothesis:

H_0: p = 0.014

(no difference) vs

H_a: p \neq 0.014

(there is a difference). The observed number of cases is 10

step 3

The test statistic for a proportion can be calculated using the formula:

z = \frac{(O - E)}{\sqrt{E(1 - p) / n}}

, where

O = 10

E = 7

p = 0.014

, and

n = 500

. Thus, we compute:

z = \frac{(10 - 7)}{\sqrt{7(1 - 0.014) / 500}}

step 4

Calculating the standard error:

SE = \sqrt{7(0.986) / 500} \approx 0.374

. Therefore,

z \approx \frac{3}{0.374} \approx 8.02

step 5

We compare the calculated

z

value to the critical

z

value for a two-tailed test at the 0.05 significance level, which is approximately ±1.96. Since 8.02 > 1.96 , we reject the null hypothesis

step 6

To find the p-value, we look up the

z

value of 8.02 in the standard normal distribution table, which gives a p-value much less than 0.0001

step 7

For part (c), we repeat the process for the girls. The expected number of cases among 300 girls is

E = 300 \times 0.01 = 3

. The observed number of cases is 4

step 8

The test statistic for the girls is calculated similarly:

z = \frac{(4 - 3)}{\sqrt{3(1 - 0.01) / 300}}

. Calculating the standard error:

SE = \sqrt{3(0.99) / 300} \approx 0.173

. Thus,

z \approx \frac{1}{0.173} \approx 5.78

step 9

Again, we compare the calculated

z

value to the critical

z

value for a two-tailed test at the 0.05 significance level. Since 5.78 > 1.96 , we reject the null hypothesis for the girls as well

step 10

The p-value for the girls' test is also much less than 0.0001

Answer

For part (a), we reject the null hypothesis indicating a significant difference in asthma incidence for boys. The p-value is less than 0.0001. For part (c), we also reject the null hypothesis for girls, with a p-value less than 0.0001.

Key Concept

Hypothesis testing for proportions allows us to determine if there is a significant difference between observed and expected values.

Explanation

The results indicate that both groups (boys and girls with smoking mothers) have a significantly higher incidence of asthma compared to the general population, as evidenced by the high test statistics and low p-values.

Iron-deficiency anemia is an important nutritional health problem in the United States. A dietary assessment was performed on 51 boys 9 to 11 years of age whose families were below the poverty level. The mean daily iron intake among these boys was found to be 12.50 mg with standard deviation 4.75 mg. Suppose the mean daily iron intake among a large population of 9- to 11-year-old boys from all income strata is 14.44 mg. We want to test whether the mean iron intake among the low-income group is different from that of the general population. (a) State the hypotheses that we can use to consider this question. (b) Carry out the hypothesis test in (a) using the critical-value method with an α level of .05, and summarize your findings. (c) What is the p-value for the test conducted in (b)

Solution by Steps

step 1

We start by stating the null and alternative hypotheses. The null hypothesis (

H_0

) states that the mean daily iron intake among the low-income group is equal to that of the general population, while the alternative hypothesis (

H_a

) states that it is different. Thus, we have:

H_0: \mu = 14.44 \quad \text{and} \quad H_a: \mu \neq 14.44

step 2

Next, we calculate the test statistic using the formula for the t-test:

t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}

where

\bar{x} = 12.50

\mu_0 = 14.44

s = 4.75

, and

n = 51

. Plugging in the values, we get:

t = \frac{12.50 - 14.44}{4.75 / \sqrt{51}}

step 3

We compute the standard error:

SE = \frac{4.75}{\sqrt{51}} \approx 0.664

and then calculate the t-value:

t \approx \frac{-1.94}{0.664} \approx -2.92

step 4

Now, we find the critical t-value for a two-tailed test at

\alpha = 0.05

with

df = n - 1 = 50

. Using a t-table, the critical values are approximately

\pm 2.009

. Since -2.92 < -2.009, we reject the null hypothesis

step 5

Finally, we summarize our findings: There is sufficient evidence to conclude that the mean daily iron intake among the low-income group is significantly different from that of the general population

step 6

To find the p-value, we look up the t-value of -2.92 in a t-distribution table or use statistical software. The p-value is approximately 0.005, which is less than 0.05, confirming our rejection of the null hypothesis

Answer

The mean daily iron intake among the low-income group is significantly different from that of the general population, with a p-value of approximately 0.005.

Key Concept

Hypothesis testing is a statistical method used to determine if there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis.

Explanation

In this case, we found that the mean iron intake among the low-income boys is significantly lower than the general population, indicating a potential health

A possible important environmental determinant of lung function in children is amount of cigarette smoking in the home. Suppose this question is studied by selecting two groups: Group 1 consists of 23 nonsmoking children 5−9 years of age, both of whose parents smoke, who have a mean forced expiratory volume (FEV) of 2.1 L and a standard deviation of 0.7 L; group 2 consists of 20 nonsmoking children of comparable age, neither of whose parents smoke, who have a mean FEV of 2.3 L and a standard deviation of 0.4 L. (a) What are the appropriate null and alternative hypotheses in this situation? (b) Carry out the test in (a) using the critical value method. (c) Provide a 95% CI for the true mean difference in FEV between 5- to 9-year-old children whose parents smoke and comparable children whose parents do not smoke.

Solution by Steps

step 1

We start by stating the null and alternative hypotheses. The null hypothesis (

H_0

) states that the mean forced expiratory volume (FEV) among children whose parents smoke is equal to that of children whose parents do not smoke, while the alternative hypothesis (

H_a

) states that it is different. Thus, we have:

H_0: \mu_1 = \mu_2 \quad \text{and} \quad H_a: \mu_1 \neq \mu_2

step 2

Next, we calculate the test statistic using the formula for the two-sample t-test:

t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}

\bar{x}_1 = 2.1, \quad s_1 = 0.7, \quad n_1 = 23, \quad \bar{x}_2 = 2.3, \quad s_2 = 0.4, \quad n_2 = 20

step 3

Plugging in the values, we find:

t = \frac{2.1 - 2.3}{\sqrt{\frac{0.7^2}{23} + \frac{0.4^2}{20}}} = \frac{-0.2}{\sqrt{\frac{0.49}{23} + \frac{0.16}{20}}}

step 4

Calculating the denominator:

\sqrt{\frac{0.49}{23} + \frac{0.16}{20}} \approx \sqrt{0.0213 + 0.008} \approx \sqrt{0.0293} \approx 0.171

step 5

Thus, the test statistic is:

t \approx \frac{-0.2}{0.171} \approx -1.17

step 6

Now, we compare the calculated t-value with the critical t-value from the t-distribution table for a two-tailed test at α = 0.05 with degrees of freedom calculated as:

df = n_1 + n_2 - 2 = 23 + 20 - 2 = 41

step 7

The critical t-value for df = 41 at α = 0.05 (two-tailed) is approximately ±2.018. Since -1.17 is within the range of -2.018 and 2.018, we fail to reject the null hypothesis

step 8

For the 95% confidence interval (CI) for the mean difference in FEV, we calculate:

CI = (\bar{x}_1 - \bar{x}_2) \pm t_{critical} \cdot SE

SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \approx 0.171

CI = (-0.2) \pm 2.018 \cdot 0.171

step 9

Calculating the margin of error:

ME \approx 2.018 \cdot 0.171 \approx 0.345

Therefore, the 95% CI is:

CI \approx (-0.2 - 0.345, -0.2 + 0.345) = (-0.545, 0.145)

Answer

The null hypothesis is not rejected, and the 95% CI for the mean difference in FEV is approximately (-0.545, 0.145).

Key Concept

Hypothesis testing and confidence intervals are used to determine if there is a significant difference between two groups.

Explanation

The analysis shows that there is no significant difference in FEV between children whose parents smoke and those whose parents do not, as indicated by the confidence interval containing zero.