CALLING ALL EXPERTS, PLEASE HELP ME WITH MY HOMEWORK!! WORTH A LOT OF POINTS AND WILL GIVE BRAINLIEST!!
2 answers:
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Given:
n = 150, sample size
Denote the sample proportion by q (normally written as
).
That is,
q = 60/150 = 0.4, sample proportion.
At the 96% confidence level, the z* multiplier is about 2.082, and the confidence interval for the population proportion is
![q \pm z^{*}[ \frac{q(1-q)}{ \sqrt{n} } ]](https://tex.z-dn.net/?f=q%20%5Cpm%20z%5E%7B%2A%7D%5B%20%5Cfrac%7Bq%281-q%29%7D%7B%20%5Csqrt%7Bn%7D%20%7D%20%5D)
That is,
0.4 +/- 2.082* √[(0.4*0.6)/150]
= 0.4 +/- 0.0833
= (0.3167, 0.4833)
= (31.7%, 48.3%)
Answer: The 96% confidence interval is about (32% to 48%)
n = 150, sample size
Denote the sample proportion by q (normally written as
That is,
q = 60/150 = 0.4, sample proportion.
At the 96% confidence level, the z* multiplier is about 2.082, and the confidence interval for the population proportion is
That is,
0.4 +/- 2.082* √[(0.4*0.6)/150]
= 0.4 +/- 0.0833
= (0.3167, 0.4833)
= (31.7%, 48.3%)
Answer: The 96% confidence interval is about (32% to 48%)