Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial consultants and services (2000 Merrill Lynch
1 answer:
Answer:
Step-by-step explanation:
Hello!
The objective is to compare the mean rating on the client satisfaction survey of two consultants A and B.
Consultant A
X₁: Rating on the client satisfaction survey given to consultant A
n₁= 16
X[bar]₁= 6.82
S₁= 0.64
Consultant B
X₂: Rating on the client satisfaction survey given to consultant B
n₂= 10
X[bar]₂= 6.25
S₂= 0.75
a. The claim is that consultant A, which is more experienced, has a higher average service rate than consultant B, which has less experience, then the hypotheses are:
H₀: μ₁ ≤ μ₂
H₁: μ₁ > μ₂
b. If both variables have a normal distribution and the population's variances are unknown but equal, the statistic for this test is a t-test for independent samples with pooled sample variance.
Sa= 0.68
c. The p-value is the probability of obtaining a value as extreme as the value of the statistic. As the test it is one-tailed and has the same direction (right), symbolically:
P(t₂₄≥2.079)= 1 - P(t₂₄<2.079)= 1 - 0.9758= 0.0242
d. The p-value:0.0242 is less than α:0.05, then the decision is to reject the null hypothesis.
Using a significance level of 5%, there is significant evidence to say that the average service rate of consultant A is higher than the average service rate of consultant B.
I hope this helps!
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Your percentile is (a hundred times) the probability that another score is less than your score. We have a normal distribution, so that probability is the integral of the standard normal from negative infinity to our normalized score.
Let's call the percentile rank
, already scaled between zero and 1.
corresponds to a z score 
because the fiftieth percentile means we got an exactly average score, 0 standard deviations away from the mean.
We know 68% of the probability will be between -1 and +1 standard deviation. So
corresponds to 
and
corresponds to 
Similarly, 95% of the probability will be between -2 and +2 standard deviations. So
corresponds to 
and
corresponds to 
That's about the list I can do off the top of my head. I think three standard deviations is 99.7%. For the rest we just consult a z table or integrated normal table. We find p in the body of the table (maybe |.5-p| depending on the table) and then the column headings tell us our z score.
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Your percentile is (a hundred times) the probability that another score is less than your score. We have a normal distribution, so that probability is the integral of the standard normal from negative infinity to our normalized score.
Let's call the percentile rank
We know 68% of the probability will be between -1 and +1 standard deviation. So
Similarly, 95% of the probability will be between -2 and +2 standard deviations. So
That's about the list I can do off the top of my head. I think three standard deviations is 99.7%. For the rest we just consult a z table or integrated normal table. We find p in the body of the table (maybe |.5-p| depending on the table) and then the column headings tell us our z score.
In this modern age, your computer can do this for you quickly