Question

Overall, approximately 45% of patients with tuberculosis (TB) have received

treatment. In a remote village consisting of 847 residents, 65 of them have been

diagnosed with TB, and among these 65 patients, 19 received treatment. Assume

significance level is 0.05. Answer the following questions:

a) The prevalence of TB in the population is 8.1%. Test the claim that prevalence in the village is different from that in the population.

b) Test the claim that the proportion of people treated with TB in the sample is different from the overall population.

Answer 1

Solution

Part (a)

Let p = prevalence proportion of TB in the population. Then, p = 0.081 (i.e., 8.1%) [given]

Let pcap = corresponding sample proportion. Then, pcap = 65/847 = 0.0767 [also given]

Claim :

Prevalence in the village is different from that in the population.

Hypotheses:

Null H₀ : p = p₀ = 0.081Vs H_A : p ≠ 0.081 [claim]

Test Statistic:

Z = (pcap - p₀)/√{p₀(1 - p₀)/n}

= (0.0767 – 0.081)/√{(0.081 x 0.919)/847}

= - 0.0043/0.009375

= - 0.4587

Distribution, Critical Value and p-value:

Under H₀, distribution of Z can be approximated by Standard Normal Distribution, provided np₀ and np₀(1 - p₀) are both greater than 10.

So, given a level of significance of α = 0.05, Critical Value = upper (2.5)% of N(0, 1), and p-value = P(Z > | Zcal |)

Using Excel Functions NORMSINV and NORMSDIST, Critical Value = 1.96 and p-value = 0.6464

Decision:

Since | Zcal | < Zcrit, or equivalently, p-value > α, H₀ is accepted.

Conclusion :

There is not enough evidence to suggest that the claim is valid. Answer

Part (b)

Let p = Proportion of people treated with TB in the population. Then, p = 0.45 (i.e., 45%) [given]

Let pcap = corresponding sample proportion. Then, pcap = 19/65 = 0.2923 [also given]

Claim :

Proportion of people treated with TB in the sample is different from that in overall population.

Hypotheses:

Null H₀ : p = p₀ = 0.45Vs H_A : p ≠ 0.45 [claim]

Test Statistic:

Z = (pcap - p₀)/√{ p₀(1 - p₀)/n}

= (0.2923 – 0.45)/√{(0.45 x 0.55)/65}

= - 0.1577/0.0617

= - 2.556

Distribution, Critical Value and p-value:

Under H₀, distribution of Z can be approximated by Standard Normal Distribution, provided np₀ and np₀(1 - p₀) are both greater than 10.

So, given a level of significance of α = 0.05, Critical Value = upper (2.5)% of N(0, 1), and p-value = P(Z > | Zcal |)

Using Excel Functions NORMSINV and NORMSDIST, Critical Value = 1.96 and p-value = 0.026

Decision:

Since | Zcal | > Zcrit, or equivalently, p-value < α, H₀ is rejected.

Conclusion :

There is enough evidence to suggest that the claim is valid. Answer

DONE