Plz answer, i will give the first person to answer brainliest! :)
2 answers:
You might be interested in
Answer:
Step-by-step explanation:
Hello!
The variable of interest is
X: time it takes a paint coating to reach 50% gloss loss (failure).
n= 8
X[bar]= 14.10hs
S= 1.67hs
Assuming X≈N(μ;σ²)
a) You have to estimate the average failure time using a 95% confidence interval. For this, considering that the sample size is small, and the population variance is unknown, you have to use the t-statistic:
[X[bar]±*
]
[14.10±2.365**]
[12.70; 15.49]
b) Interpretation:
Using a 95% confidence level you'd expect that the interval [12.70; 15.49]hours contains the true mean failure time for the paint coating.
c)
Using the estimates of the mean and standard deviation based on the sample of 8 panes as the true mean and standard deviation of failure time of a panel (assumed to be <u>normally distributed</u>), find the cut-off failure time that separates, among all the panels, the bottom 10% from the rest, i.e. the 10th percentile.
μ= 14.10
σ= 1.67
You have to find a value of the variable X that has below it 10% of the distribution, symbolically:
P(X ≤ x₀)= 0.10
To do so, you have to work using the tabulated standard normal distribution (Z) and then, using the population values μ and δ, "transform" the value of Z to a value of X. (i.e. you have to reverse the standardization to reach the corresponding value of X)
Using the Z-distribution Z= (X-μ)/σ
P(Z ≤ z₀)= 0.10
Using the left entry of the table, since we are working in the left tail of the distribution.
z₀= -1.282
Now using the formula you have to calculate the value of X
z₀= (x₀-μ)/σ
z₀*σ= x₀-μ
(z₀*σ)+μ= x₀
x₀= (z₀*σ)+μ= (-1.282*1.67) + 14.10= 11.95906 ≅ 11.96hours
The value that separates the bottom 10% of the panels from the rest is 11.96 hours.
I hope it helps!