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Solution:

Given: the foundation would like to know if the pregnancy ratefor teen girls who participate in the educational programing atTPPC is significantly lower than the citywide average (currently21%).

That is we have to test if proportion the pregnancy rate forteen girls who participate in the educational programing at TPPC issignificantly lower than the citywide average (currently 21%).

That is: p < 0.21

Sample size = n = 65

x = number of girls who completed the Coalitions programming andcontacted went on to become pregnant = 8

Thus sample proportion:

Thus we use following steps:

Step 1) State H0 and H1:

H0: p =0.21          Vs       H1: p < 0.21

Step 2) Find test statistic

We use z test statistic for proportion.

Step 3) z test statistic value:

Since level of significance is not given, we use level ofsignificance =

Thus look in z table for Area = 0.0500 or its closest area andfind z value

Area 0.0500 is in between 0.0495 and 0.0505 and both the areaare 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 = -1.645

Step 4) Decision rule:

Reject H0, if z test statistic value < Z_critical =-1.645, otherwise we fail to reject H0.

Since z test statistic value = -1.72 < Z_critical =-1.645, we reject H0.

Step 5) Conclusion:

Since we have rejected H0, we conclude that: the pregnancy ratefor teen girls who participate in the educational programing atTPPC is significantly lower than the citywide average ( that is: p< 21%).

Since the director has proved that the pregnancy rate for teengirls who participate in the educational programing at TPPC issignificantly lower than the citywide average ( that is: p <21%), the director should not be concerned about presenting theirfindings to the foundation.