A bus company has contracted with a local high school to carry 450 students on a field trip. The company has 18 large buses whic
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Answer:
Given:
Mean, u = 135
Sample size, n = 42
Sample mean, x' = 126
Standard deviation = 16
Significance level = 0.05
1) The null and alternative hypotheses:
H0 : u = 135 (the revenue per day is $135)
H1 : u < 135 (the revenue per day is less than $135)
2) In this case, we have a left tailed test.
Let's use the formula:
t = - 3.645
Critical value: at a significance level of 0.05, left tailed test,
tcritical = -1.683
We reject null hypothesis, H0, since t calculated, -3.645 is less than tcritical, -1.683.
3) Given, a = 0.05, df = 41, left tailed test,
t-value = 2.02
For the corresponding confidence interval, we have:
CI = (121.014, 130.986)
4) Conclusion:
We reject the estimated potential revenue advised by the consultant, H0, as it is too high, because the t-calculated falls in rejection region(i.e, less than critical value), also the upper limit of the confidence interval is less than $135.
Answer:
a) p + m = 10 and
b) The mixing amount is 5 lb peanut and 5 lb mixture.
Step-by-step explanation:
Given that p pounds of peanuts and m pounds of the 80% almond and 20% peanut mixture are used to make 10 pounds of 60% peanuts and 40% of almonds mixture.
Now, we can write that p + m = 10 ........ (1)
Now, m pounds of 80% almond and 20% peanut mixture contain 0.2m pounds of peanuts and 0.8m pounds of almonds.
Now, from the condition given it can be written that
.......... (2)
⇒
⇒ 2p + 0.4m = 2.4m
⇒ 2p = 2m
⇒ p = m
Now from equation (1) we get p = m = 5 pounds.
a) Therefore, equations (1) and (2) are the system of equations that models the situation.
b) The mixing amount is 5 lb peanut and 5 lb mixture. (Answer)