In: Statistics and Probability
Here, we have given that,
n1= number of healthy boys aged 6-11 = 546
x1= number of healthy boys aged 6-11 found that they were overweight=74
=Sample proportion of healthy boys aged 6-11 found that they were overweight
=
n2= number of healthy girls aged 6-11 = 455
x2= number of healthy girls aged 6-11 found that they were overweight=75
=Sample proportion of healthy girls aged 6-11 found that they were overweight
=
Claim: To check whether the population proportion of healthy boys aged 6-11 found that they were overweight is less than the population proportion of healthy girls aged 6-11 found that they were overweight.
The null and alternative hypotheses are as follows:
Versus
Where p1 = The population proportion of healthy boys aged 6-11 found that they were overweight.
p2= The population proportion of healthy girls aged 6-11 found that they were overweight.
This is the left one-tailed test.
Here, we are using the two-sample proportion test.
Now, we can find the test statistics
Z-statistics=
=
=
= -1.29
The test statistics is -1.29.
Now, we can find the p-value
P-value =P( Z < -1.29) as this is left one tailed test
= 0.0985 Using standard normal z table see the value corresponding to the z=-1.29
we get the p-value is 0.0985
Decision:
= level of significance= 0.10
Here, P-value (0.0985) less than (>) 0.10
Conclusion:
we reject the Null hypothesis Ho
we conclude that there is sufficient evidence to support the claim the population proportion of healthy boys aged 6-11 found that they were overweight is less than the population proportion of healthy girls aged 6-11 found that they were overweight.i.e. .