Question

A community health service initiates a cardiovascular risk screening intervention as one of the activities in the overall health promotion goal of reducing cardiovascular risk for men aged 40 or over in the local community. In the planning process, an impact indicator for this intervention is established: 60 per cent or more of at risk men in the population will consult a general practitioner (GP) to reduce their level of risk ─ meaning that if the intervention is successful, 60% or more at risk men will consult GP.

A sample of 140 participants in the program who were screened as at risk are followed up by telephone and interviewed to determine whether they had consulted a GP. The survey results indicated that 89 of those screened as at risk had consulted a GP about their screening result. Did the intervention achieved desired impact for the at risk participants of the intervention? Let α=0.05.

Answer 1

Solution:

   We are given that: In the planning process, an impact indicator for this intervention is established: 60 per cent or more of at risk men in the population will consult a general practitioner (GP) to reduce their level of risk.

   A sample of 140 participants in the program who were screened as at risk are followed up by telephone and interviewed to determine whether they had consulted a GP. The survey results indicated that 89 of those screened as at risk had consulted a GP about their screening result.

That is:

Sample size = n = 140

x = number of participants screened as at risk had consulted a GP about their screening result =89

Then sample proportion =

We have to test if the intervention achieved desired impact for the at risk participants of the intervention.

That is we have to test if 60% or more at risk men will consult GP.

Step 1) State H0 and H1:

Vs

Step 2) Test statistic:

Step 3) Find critical value:

This is left tailed test , hence look in z table for Level of significance = 0.05 area and find z value.

Area 0.0500 is in between 0.0495 and 0.0505 and both the area are at same distance from 0.0500

Thus we look for both area and find both z values

Thus Area 0.0495 corresponds to -1.65 and 0.0505 corresponds to -1.64

Thus average of both z values is : ( -1.64+ - 1.65) / 2 = -1.645

Thus z critical value = -1.645

Step 4) Decision rule: Reject H0 , if z test statistic value < z critical valuel otherwise we fail to reject H0.

Since z test statistic value = 0.86 > z critical value = -1.645 , we fail to reject H0.

Step 5) Conclusion:

Since we failed to reject H0, that is we retain null hypothesis , which indicates that 60% or more at risk men will consult GP. Thus the intervention achieved desired impact for the at risk participants of the intervention.