✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧
Hello!
✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧
❖ 2. a: Associative property b: Associative property c: Associative property
d: Associative property
~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡
~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ
The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]
<h3>
How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>
Suppose that we have:
- Sample size n > 30
- Sample mean =
- Sample standard deviation = s
- Population standard deviation =
- Level of significance =
Then the confidence interval is obtained as
- Case 1: Population standard deviation is known
- Case 2: Population standard deviation is unknown.
For this case, we're given that:
- Sample size n = 90 > 30
- Sample mean = = 138
- Sample standard deviation = s = 34
- Level of significance = = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).
At this level of significance, the critical value of Z is: 
= ±1.645
Thus, we get:
![CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]](https://tex.z-dn.net/?f=CI%20%3D%20%5Coverline%7Bx%7D%20%5Cpm%20Z_%7B%5Calpha%20%2F2%7D%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5CCI%20%3D%20138%20%5Cpm%201.645%5Ctimes%20%5Cdfrac%7B34%7D%7B%5Csqrt%7B90%7D%7D%5C%5C%5C%5CCI%20%5Capprox%20138%20%5Cpm%205.896%5C%5CCI%20%5Capprox%20%5B138%20-%205.896%2C%20138%20%2B%205.896%5D%5C%5CCI%20%5Capprox%20%5B132.104%2C%20143.896%5D%20%5Capprox%20%5B130.10%2C%20143.90%5D)
Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]
Learn more about confidence interval for population mean from large samples here:
brainly.com/question/13770164
Answer:
x = 4/3
Step-by-step explanation:
15 - 5x = 6x + 3 - 2x
In this kind of equation, we try to seperate the x's and the numbers.
So we add 5x to both sides and subtract 3 from both sides to seperate these values
15 - 3 - 5x + 5x = 6x + 3 - 3 - 2x + 5x
=> 12 = 9x
=> 12/9 = x
=> 4/3 = x
=> x = 4/3
To check, we can plug this in.
15 - 5(4/3) = 6(4/3) + 3 - 2(4/3)
15 - 20/3 = 24/3 + 3 - 8/3
45/3 - 20/3 = 24/3 + 9/3 - 8/3
25/3 = 25/3
So our answer works.
Answer:
Kindly check explanation
Step-by-step explanation:
H0 : μ = 5500
H1 : μ > 5500
The test statistic assume normal distribution :
Test statistic :
(Xbar - μ) ÷ s/sqrt(n)
(5625.1 - 5500) ÷ 226.1/sqrt(15) = 2.1429 = 2.143
Pvalue from test statistic :
The Pvalue obtained using the calculator at df = 15 - 1 = 14 is 0.025083
α = 0.05
Since ;
Pvalue < α
0.025083 < 0.05 ; Reject H0
The confidence interval :
Xbar ± Tcritical * s/sqrt(n)
Tcritical at 95% = 1.761 ;
margin of error = 1.761 * 226.1/sqrt(15) = 102.805
Lower boundary : (5625.1 - 102.805) = 5522.295
(5522.295 ; ∞)
The hypothesized mean does not occur within the constructed confidence boundary. HENCE, there is significant eveidbce to support the claim that the true mean life of a biomedical device is greater than 5500
1/4 of 80 is 20, so they have 60 left aftef yesterday. then sold 3/5 of 60 which is 36. so the book store has 24 left
