Can you help i need to finish this by today!
1 answer:
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Answer:
At the 0.05 level of significance, the claim made by Blue Diamond is true
Step-by-step explanation:
Given that the Blue Diamond Company advertises that their nut mix contains (by weight) 40% cashews, 15% Brazil nuts, 20% almonds and only 25% peanuts.
(Two tailed chi square test at 5% significance level)
Chi square is calculated as (Obs_exp)^2/Expected
Cashews Brazil nuts Almonds Peanuts Total
Observed 5 6 6 3 20
Expected 40% 15% 20% 25% 1
Expected 8 3 4 5 20
Chi square 1.125 3 1 0.8 5.925
df = 3
p value = 0.1153
Since p value >0.0 we accept null hypothesis.
Answer:
A task time of 177.125s qualify individuals for such training.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by
After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem, we have that:
A distribution that can be approximated by a normal distribution with a mean value of 145 sec and a standard deviation of 25 sec, so .
The fastest 10% are to be given advanced training. What task times qualify individuals for such training?
This is the value of X when Z has a pvalue of 0.90.
Z has a pvalue of 0.90 when it is between 1.28 and 1.29. So we want to find X when .
So
A task time of 177.125s qualify individuals for such training.