Complete question is:
A hospital claims that the proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%. In a random sample of 170 babies born in this hospital, 56 weighed over 7 pounds. Is there enough evidence to reject the hospital's claim at the level of significance?
Answer:
Yes, there is enough evidence to reject the claim.
Step-by-step explanation:
We are given;
n = 170
x = 56
So, will use one sample proportion test to solve this.
p^ = x/n
p^ = 56/170
p^ = 0.3294
Since the proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%.
Thus;
Null Hypothesis H0: p ≠ 0.36
Alternative Hypothesis Ha: p = 0.36
Formula for test statistic = (p^ - p)/√(p(1 - p)/n)
This gives;
Test statistic = (0.3294 - 0.36)/√(0.36(1 - 0.36)/170)
Test statistic = -0.8311
From z-table and online z-calculator, the p - value is 0.203.
level of significance is; α = 0.05
Now, Since the p value < α, we reject the null hypothesis .
Thus, the claim is true
