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Julli [10]
2 years ago
12

Assume that a sample is used to estimate a population proportion p. Find the 99.5% confidence interval for a sample of size 385

with 70.9% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.
Mathematics
1 answer:
LuckyWell [14K]2 years ago
7 0

Answer: 0.644

Step-by-step explanation:

The confidence interval for population proportion (p) is given by :-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

Given : Significance level : \alpha: 1-0.995=0.005

Critical value : z_{\alpha/2}=z_{0.0025}= 2.80

Sample size : n= 385

Sample proportion : \hat{p}=0.709

Then , the 99.5% confidence interval for population proportion is given by :-

0.709\pm (2.80)\sqrt{\dfrac{0.709(1-0.709)}{385}}\\\\\approx0.709\pm0.065\\\\=(0.709-0.065,0.709+0.065)=(0.644,0.774)

Hence, the 99.5% confidence interval for population proportion :

0.644

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3 0
2 years ago
Anyone know how to do these and explain so I can do the rest lol.
klasskru [66]

Answer:

Option B. m⁷/n²  is the correct answer.

Step-by-step explanation:

Identity,

xᵃ * xᵇ = xᵃ⁺ᵇ  

xᵃ/xᵇ = xᵃ⁻ᵇ

x⁻ᵃ = 1/xᵃ

It is given that,

m⁻⁶ n⁻³/m⁻¹³ n⁻¹

Using  these three identities we can write,

m⁻⁶ n⁻³/m⁻¹³ n⁻¹ = m⁻⁶ m¹³/n⁻¹n³   (since x⁻ᵃ = 1/xᵃ)

m⁻⁶ m¹³/n⁻¹n³ =m⁽⁻⁶⁺¹³⁾/n⁽⁻¹⁺³⁾ = m⁷/n²  (since xᵃ * xᵇ = xᵃ⁺ᵇ   and xᵃ/xᵇ = xᵃ⁻ᵇ)

Therefore Option B. m⁷/n² is the correct answer

6 0
3 years ago
Read 2 more answers
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