Answer:
a) Sample size = 1691
b) 95% Confidence Interval = (0.3696, 0.4304)
Explanation:
(a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?
The margin of error is given by
Where z is the corresponding z-score for 90% confidence level
z = 1.645 (from z-table)
for p = 0.50 and 2% margin of error, the required sample size would be
(b) The advocacy group took a random sample of 1000 consumers who recently purchased this mobile phone and found that 400 were happy with their purchase. Find a 95% confidence interval for p.
The sample proportion is
p = 400/1000
p = 0.40
z = 1.96 (from z-table)
n = 1000
The confidence interval is given by
Therefore, we are 95% confident that the proportion of consumers who bought the newest generation of mobile phone were happy with their purchase is within the range of (0.3696, 0.4304)
What is Confidence Interval?
The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation:
Answer:
Pineapple packs vitamin B, fiber, and a heaping dose of vitamin C, after all. But at the end of the day, maybe it's okay if your va jay jay just smells and tastes… like a va jay jay. Your scent naturally changes a bit throughout your menstrual cycle, but it probably shouldn't ever taste or smell like a tropical fruit.
Explanation:
Answer:
the maximum length of specimen before deformation is found to be 235.6 mm
Explanation:
First, we need to find the stress on the cylinder.
Stress = σ = P/A
where,
P = Load = 2000 N
A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4
A = 1.0752 x 10^-5 m²
σ = 2000 N/1.0752 x 10^-5 m²
σ = 186 MPa
Now, we find the strain (∈):
Elastic Modulus = Stress / Strain
E = σ / ∈
∈ = σ / E
∈ = 186 x 10^6 Pa/107 x 10^9 Pa
∈ = 1.74 x 10^-3 mm/mm
Now, we find the original length.
∈ = Elongation/Original Length
Original Length = Elongation/∈
Original Length = 0.41 mm/1.74 x 10^-3
<u>Original Length = 235.6 mm</u>
1.5 nautical miles per 1,000 feet
