A hospital claims that the proportion, , of full-term babies born in their hospital that weigh more than pounds is . In a random sample of babies born in this hospital, weighed over pounds. Is there enough evidence to reject the hospital's claim at the

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?

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

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743

5+2x=a

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