At a pet store, Davina counted 12 parrots out of 20 birds. Which is an equivalent ratio of parrots to birds at the pet store?
2 answers:
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Answer:
a) No
b) No
c) P value is more than 0.05.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 12 ounces
Sample size, n = 49
P-value = 0.136
First, we design the null and the alternate hypothesis
We use One-tailed z test to perform this hypothesis.
a) Alpha, α = 0.05
Since, p-value > α,
The null hypothesis should not be rejected. We accept the null hypothesis and reject the alternate hypothesis. We conclude that the average chip weight is 12 ounces per bag.
b) Alpha, α = 0.10
Since, p-value > α,
The null hypothesis should not be rejected. We accept the null hypothesis and reject the alternate hypothesis. We conclude that the average chip weight is 12 ounces per bag.
c) The evidence is statistically significant at the .05 level means that the p value is more than 0.05.
Answer:
a)
Operating room x 1 2 3 4
P(x) 0.15 0.25 0.4 0.2
b)
The graph of probability distribution is in attached image.
c)
All the probabilities lies in the range (0 to 1) and the probabilities add up to 1 so, the computed probability distribution is the valid probability distribution.
Step-by-step explanation:
Number of operating rooms Days
1 3
2 5
3 8
4 4
a)
Sum of frequencies=1+2+3+4=10
Operating rooms x frequency f Relative frequency
1 3 3/20=0.15
2 5 5/20=0.25
3 8 8/20=0.4
4 4 4/20=0.2
So, using the relative frequency approach, a discrete probability distribution for number of operating rooms in use on any given day is
Operating room x 1 2 3 4
P(x) 0.15 0.25 0.4 0.2
b)
The graph of probability distribution is in attached image.
c)
There are two conditions for a probability distribution to be valid
1. All probability must ranges from 0 to 1.
2. All probabilities must add up to 1.
We can see that all the probabilities lies in the range (0 to 1), so, condition 1 is satisfied.
For condition 2,
sum[p(x)]= 0.15+0.25+0.4+0.2=0.4+0.6=1.
As the probabilities add up to 1, so the condition 2 is also satisfied.
Thus, the computed probability distribution is the valid probability distribution.
