Mathematics, 27.11.2019 02:31, choudharykaran7997

Periodically, merrill lynch customers are asked to evaluate merrill lynch financial consultants
and services. higher ratings on the client satisfaction survey indicate better service,
with 7 the maximum service rating. independent samples of service ratings for two
financial consultants are summarized here. consultant a has 10 years of experience,
whereas consultant b has 1 year of experience. use α= .05 and test to see whether the
consultant with more experience has the higher population mean service rating.

consultant an1=16x¯¯¯1=6.82s1=.64 consultant bn2=10x¯¯¯2=6.25s2=.75

a. state the null and alternative hypotheses.
b. compute the value of the test statistic.
c. what is the p-value?
d. what is your conclusion?

Answers: 2

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Answers

answered: liluzisquirt7378

a) Null hypothesis: $\mu_{A} \leq \mu_{B}$

Alternative hypothesis: $\mu_{A} \mu_{B}$

b) $t=\frac{6.82-6.25}{\sqrt{\frac{0.64^2}{16}+\frac{0.75^2}{10}}}}=1.992$

c) $p_v =P(t_{(24)}1.992)=0.0289$

d) If we compare the p value and the significance level given $\alpha=0.05$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the more experience consultant A have a significant higher rate compared to the consultant B with less experience at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{A}=6.82$ represent the mean for the sample of Consultant A

$\bar X_{B}=6.25$ represent the mean for the sample of Consultant B

$s_{A}=0.64$ represent the sample standard deviation for the sample of Consultant A

$s_{B}=0.75$ represent the sample standard deviation for the sample of bonsultant B

$n_{A}=16$ sample size selected for the Consultant A

$n_{B}=10$ sample size selected for the Consultant B

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the Consultant A (more experience) is higher than the mean for the Consultant B, the system of hypothesis would be:

Null hypothesis: $\mu_{A} \leq \mu_{B}$

Alternative hypothesis: $\mu_{A} \mu_{B}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{stick}-\bar X_{Liquid}}{\sqrt{\frac{s^2_{stick}}{n_{stick}}+\frac{s^2_{Liquid}}{n_{Liquid}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{6.82-6.25}{\sqrt{\frac{0.64^2}{16}+\frac{0.75^2}{10}}}}=1.992$

Part c: P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{A}+n_{B}-2=16+10-2=24$

Since is a one side test the p value would be:

$p_v =P(t_{(24)}1.992)=0.0289$

Part d: Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the more experience consultant A have a significant higher rate compared to the consultant B with less experience at 5% of significance.

answered: Guest

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column