Answer:
Step-by-step explanation:
Hello!
Your study variable is X: "number of ColorSmart-5000 that didn't need repairs after 5 years of use, in a sample of 390"
X~Bi (n;ρ)
ρ: population proportion of ColorSmart-5000 that didn't need repairs after 5 years of use. ρ = 0.95
n= 390
x= 303
sample proportion ^ρ: x/n = 303/390 = 0.776 ≅ 0.78
Applying the Central Limit Theorem you approximate the distribution of the sample proportion to normal to obtain the statistic to use.
You are asked to estimate the population proportion of televisions that didn't require repairs with a confidence interval, the formula is:
^ρ±
* √[(^ρ(1-^ρ))/n]
= 
= 2.58
0.78±2.58* √[(0.78(1-0.78))/390]
0.0541
[0.726;0.834]
With a confidence level of 99% you'd expect that the interval [0.726;0.834] contains the true value of the proportion of ColorSmart-5000 that didn't need repairs after 5 years of use.
I hope it helps!
Hey there!
-1 ¼ • 9
= - (4+1)/4 • 9
= -5/4 • 9
= -(5 × 9)/4
= <u>-</u><u> </u><u>4</u><u>5/4</u>
= <u>- 11.25</u>
Hope it helps ya!
Answer:
Parallel
Step-by-step explanation:
-x+6 slope:-1
X+y=20 slope -1
Hence the answer is parallel
If inspection department wants to estimate the mean amount with 95% confidence level with standard deviation 0.05 then it needed a sample size of 97.
Given 95% confidence level, standard deviation=0.05.
We know that margin of error is the range of values below and above the sample statistic in a confidence interval.
We assume that the values follow normal distribution. Normal distribution is a probability that is symmetric about the mean showing the data near the mean are more frequent in occurence than data far from mean.
We know that margin of error for a confidence interval is given by:
Me=
α=1-0.95=0.05
α/2=0.025
z with α/2=1.96 (using normal distribution table)
Solving for n using formula of margin of error.
n=
=96.4
By rounding off we will get 97.
Hence the sample size required will be 97.
Learn more about standard deviation at brainly.com/question/475676
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The given question is incomplete and the full question is as under:
If the inspection division of a county weights and measures department wants to estimate the mean amount of soft drink fill in 2 liters bottles to within (0.01 liter with 95% confidence and also assumes that standard deviation is 0.05 liter. What is the sample size needed?
Answer:
None! .7632 is smaller than 9.
Step-by-step explanation:
