Answer:
Claim :The proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%.
n = 170
x = 56
We will use one sample proportion test
![\widehat{p}=\frac{x}{n}](https://tex.z-dn.net/?f=%5Cwidehat%7Bp%7D%3D%5Cfrac%7Bx%7D%7Bn%7D)
![\widehat{p}=\frac{56}{170}](https://tex.z-dn.net/?f=%5Cwidehat%7Bp%7D%3D%5Cfrac%7B56%7D%7B170%7D)
![\widehat{p}=0.3294](https://tex.z-dn.net/?f=%5Cwidehat%7Bp%7D%3D0.3294)
The proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%.
![H_0:p \neq 0.36 \\H_a:p= 0.36](https://tex.z-dn.net/?f=H_0%3Ap%20%5Cneq%200.36%20%5C%5CH_a%3Ap%3D%200.36)
Formula of test statistic =![\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cwidehat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D)
![=\frac{0.3294-0.36}{\sqrt{\frac{0.36(1-0.36)}{170}}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.3294-0.36%7D%7B%5Csqrt%7B%5Cfrac%7B0.36%281-0.36%29%7D%7B170%7D%7D%7D)
=−0.8311
Now refer the p value from the z table
P-Value is .202987 (Calculated by online calculator)
Level of significance α = 0.05
Since p value < α
So we reject the null hypothesis .
Hence the claim is true
Answer:
The question is not explained well, Please explain it better or take a screenshot or something
Step-by-step explanation:
Answer:
8863
Step-by-step explanation:
8.863 X 1,000=8863
Answer:
Step-by-step explanation:
eq. of circle with center (h,k) and radius=r is
(x-h)²+(y-k)²=r²
reqd. eq. is
(x-2)²+(y+5)²=9²