Assume that 6 people, including. the husband and wife pair, apply for 5 sales positions. People are hired at random. What is the
1 answer:
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An interesting twist to a binomial distribution problem.
Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899 (from context, it is >= and not =, which is almost impossible)
From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.
The number of successes, x, is then given by
where
p=probability of success
n=number of trials
x=number of successes
Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find
Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.
Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368
Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more)
=1-S
=0.898005, missing target by 0.1%
So she will have to make 11 phone calls, bring up the probability to 93.9%. The work is similar to that of n=10.
Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899 (from context, it is >= and not =, which is almost impossible)
From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.
The number of successes, x, is then given by
p=probability of success
n=number of trials
x=number of successes
Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find
Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.
Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368
Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more)
=1-S
=0.898005, missing target by 0.1%
So she will have to make 11 phone calls, bring up the probability to 93.9%. The work is similar to that of n=10.
Answer:
Part A
b. 14.6 ± 7.38
Part B
b. 3.43
Part C
a. P-value < 0.01
Part D
b. There is sufficient evidence to reject the null hypothesis
Step-by-step explanation:
Part A
The given data are;
The number of seedlings in the field = 20
The number of seedlings selected to receive herbicide A = 10
The number of seedlings selected to receive herbicide B = 10
The height in centimeters of seedlings treated with Herbicide A, = 94.5 cm
The standard deviation, s₁ = 10 cm
The height in centimeters of seedlings treated with Herbicide B, = 109.1 cm
The standard deviation, s₂ = 9 cm
The 90% confidence interval for μ₂ - μ₁, is given as follows;
The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;
The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18
α = 100% - 90% = 10%
∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05
= 1.734
C.I. ≈ 14.6 ± 7.37714603353
The 90% C.I. ≈ 14.6 ± 7.38
b. 14.6 ± 7.38
Part B
With the hypotheses are given as follows;
H₀; μ₂ - μ₁ = 0
Hₐ; μ₂ - μ₁ ≠ 0
The two sample t-statistic is given as follows;
The two sample t-statistic ≈ 3.43
b. 3.43
Part C
From the t-table, the p-value, we have, the p-value < 0.01
a. P-value < 0.01
Part D
Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis
b. There is sufficient evidence to reject the null hypothesis.