A pharmaceutical company sells bottles of 500 calcium tablets in two dosages: 250 milligram and 500 milligram. Last month, the c
2 answers:
Answer:
The price of 250mg bottle is $8 and the price of 500mg bottle is $12.
Step-by-step explanation:
The company sold last month 250 mg tablets = 2,200 bottles
Bottle of 500 mg tablets = 1,800 bottles
Total sales revenue = $39,200
This month sales team has targeted sales of $44,000 for this month.
They have target of 2,200 bottles of 250 mg tablets that is same as last month.
and 2,200 bottles of 500 mg tablets = 2,200 - 1,800 = 400 more bottles
Now we calculate the price of 500mg tablets by subtracting last month's revenue from this month's target.
price of 400 bottles of 500mg tablets = 44,000 - 39,200 = $4,800
Therefore, the price of one bottle of 500 mg tablet is = = $12 (one bottle of 500mg)
Last month they sold 1,800 bottles, so the revenue generated by 500mg bottles = 1,800 × 12 = $21,600
and revenue generated by 2200 bottles of 250mg tablets = 39,200 - 21,600 = $17,600
Therefore, the price of one bottle of 250mg tablets = 17,600 ÷ 2,200 = $8
The price of a 250mg bottle is $8 and the price of a 500mg bottle is $12.
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Answer:
Step-by-step explanation:
Hello!
The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.
Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.
Be:
X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)
n₁= 12
X[bar]₁= 20.50 hs
S₁²= 199.76hs² (S₁= 14.13hs)
X₂: Battery duration between charges when the primary usage of the phone is internet applications.
n₂= 10
X[bar]₂= 8.50
S₂²= 33.29hs² (S₂= 5.77hs)
Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)
The parameters of interest are σ₁² and σ₂²
a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1
The hypotheses are:
H₀: σ₁²/σ₂² ≤ 1
H₁: σ₁²/σ₂² > 1
There is no level of significance listed so I've chosen α: 0.05
b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:
S²= *[∑X²-(∑X)²/n]
For the first sample:
n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36
S₁²= *[7240.36-(246)²/12]= 199.76hs²
S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs
For the second sample:
n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12
S₂²= *[1022.12-(85)²/10]= 33.2911hs²
S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs
c)
For this hypothesis test, the statistic to use is a Snedecors F:
This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:
The rejection region is then F ≥ 3.10
p-value: 0.006
Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.
Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.
I hope this helps!
