A sample of size 20 from a normal distribution has sample mean 3.5 and sample variance 14.75. Find 90% confidence intervals for
1 answer:
Answer:
Variance =9.2983, 27.6927
Standard deviation = 3.0493, 5.2624
Step-by-step explanation:
n = 20
S = 14.75
X² = chisquare
N-1 = degree of freedom
Alpha = 0.1
Alpha/2 = 0.1/2 = 0.05
1-0.1/2 = 0.95
Using statistics table
X² 19,0.05 = 30.14
X² 19, 0.95 = 10.12
A. 19(14.75)/30.14 , 19(14 75)/10.12
= 9.2983, 27.6927 for the variance of the population
B. Standard deviation = square root of variance
= √9.2983, √27.6927
= 3.0493, 5.2624
Please use attachment for better understanding
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Answer:
- Part A: The price of fuel A is decreasing by 12% per month.
- Part B: Fuel A recorded a greater percentage change in price over the previous month.
Explanation:
<u>Part A:</u>
The function calculates the price of fuel A each month by multiplying the price of the month before by 0.88.
Month price, f(x)
1 2.27 (0.88) = 1.9976 ≈ 2.00
2 2.27(0.88)² = 1.59808 ≈ 1.60
3 2.27(0.88)³ = 1.46063 ≈ 1.46
Then, the price of fuel A is decreasing.
The percentage per month is (1 - 0.88) × 100 = 12%, i.e. the price decreasing by 12% per month.
<u>Part B.</u>
<u>Table:</u>
m price, g(m)
1 3.44
2 3.30
3 3.17
4 3.04
To find if the function decreases with a constant ration divide each pair con consecutive prices:
- ratio = 3.30 / 3. 44 = 0.959 ≈ 0.96
- ratio = 3.17 / 3.30 = 0.960 ≈ 0.96
- ratio = 3.04 / 3.17 = 0.959 ≈ 0.96
Thus, the price of fuel B is decreasing by (1 - 0.96) × 100 =4%.
Hence, the fuel A recorded a greater percentage change in price over the previous month.