Based on a random sample of 25 units of product X, the average weight is 102 lbs., and the sample standard deviation is 10 lbs.
1 answer:
Answer:
Step-by-step explanation:
Given that for a sample size of 25, sample mean = 102 lbs. and sample std deviation = 10 lbs.
(Right tailed test)
Given that we assume population is normally distributed
Test statistic:
Mean difference = +2
Std error of sample =
t = test statistic = Mean diff/Std error
degree of freedom
Critical value of t = 1.711
Our test statistic >1.71
Accept alternate hypothesis
Because there is significant difference between the test statistic and test statistic lies above critical value.
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Answer:
Step-by-step explanation:
Answer:
a. The amount that is saved at the expiration of the 5 year period is $22,769.20¢
b. The amount of interest is $2,769.20¢
Step-by-step explanation:
Since the amount that is deposited every year for a period of five years is $4,000 and the rate of the interest is 6.5%. We can always calculate the amount that is saved at the expiration of the five years.
We will first state the formula for calculating the future value of annuity:-
Future value of annuity =
Where P is the amount deposited per year.
r is the rate of interest
t is the time or period
and in this case, the actual value of P = $4,000
rate of interest, r is 6.5% = 0.065
time, t is 5 years.
Substituting e, we have:
Fv of annuity =
= 4,000 × [((1.065)^5)- 1/0.065]
= 4,000 × [(1.37 - 1)/0.065]
= 4,000 × (0.37/0.065)
= 4,000 × 5.6923
= $22,769.20¢
a. Therefore the amount that is saved at the end of the five (5) years is $22,769.20¢
b. To find the interest, we will calculate the amount of deposit made during the period of five years and subtract the sum from the current amount that is saved ($22,769.29¢).
Since I deposited 4,000 every year for five years, the total amount of deposit I made at the period =
4,000 × 5 = $20,000
The amount of interest is then = $22,769.20¢ - $20,000 = $2,769.20¢