/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Periodically, Merrill Lynch cust... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\mathrm{B}\) has 1 year of experience. Use \(\alpha=.05\) and test to see whether the consultant with more experience has the higher population mean service rating. Consultant A \(\quad\) Consultant B\\[\begin{array}{ll} n_{1}=16 & n_{2}=10 \\\\\bar{x}_{1}=6.82 & \bar{x}_{2}=6.25 \\\s_{1}=.64 & s_{2}=.75\end{array}\\] 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?

Short Answer

Expert verified
Consultant A is suggested to have a higher mean service rating (p-value 0.015 < 0.05).

Step by step solution

01

State the Hypotheses

For this problem, we need to test if Consultant A (with more experience) has a higher population mean service rating than Consultant B. The null hypothesis \((H_0)\) is that the mean rating of Consultant A \(\mu_1\) is equal to or less than that of Consultant B \(\mu_2\), which can be stated as \(H_0: \mu_1 \leq \mu_2\). The alternative hypothesis \((H_a)\) is that the mean rating of Consultant A is greater, so \(H_a: \mu_1 > \mu_2\).
02

Compute the Test Statistic

For the given exercise, we should use a two-sample t-test for the difference between means. The test statistic is computed using: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Plugging in the given values, we get \( t = \frac{6.82 - 6.25}{\sqrt{\frac{0.64^2}{16} + \frac{0.75^2}{10}}} \). After calculating, you find \(t \approx 2.34\).
03

Determine Degrees of Freedom

Before finding the p-value, determine the degrees of freedom using the formula: \[ df = \left( \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \right) \] After calculating, the degrees of freedom is approximately 19.5. We round it to 19 for practical purposes.
04

Find the P-value

We use the t-distribution table or statistical software to find the p-value for \(t = 2.34\) with 19 degrees of freedom. The p-value is approximately 0.015, when referring to standard t-tables. This value is found for a one-tailed test as per our alternative hypothesis.
05

Make a Conclusion

Since the p-value \(0.015\) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis \(H_0\). This means that there is sufficient evidence to suggest that the consultant with more experience has a higher population mean service rating than the consultant with less experience.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In this case, we are comparing the service ratings of two financial consultants with different levels of experience. The main purpose is to evaluate whether the observed difference in means is due to a true difference in the population or just by random chance. For this, we employ the two-sample t-test, which is suitable when comparing two independent samples. The calculation involves the formula:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] - \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means,- \(s_1\) and \(s_2\) are the sample standard deviations,- \(n_1\) and \(n_2\) are the sample sizes of each group.This test calculates the probability that any observed difference between the groups is due purely to sampling variability. If this probability (captured by the p-value) is low, we can conclude that the observed difference is statistically significant.
p-value
The p-value is critical in hypothesis testing as it helps determine the strength of the results. It answers the question: "What is the probability of observing the data, or something more extreme, if the null hypothesis is true?" In our scenario, the null hypothesis states that there is no difference in the service ratings between the more and less experienced consultants. - A low p-value (typically \(<0.05\)) indicates strong evidence against the null hypothesis, suggesting that it is unlikely the difference in means occurred by chance. - Conversely, a high p-value suggests insufficient evidence to reject the null hypothesis.For our exercise, a calculated p-value of 0.015 was obtained for the t-statistic with 19 degrees of freedom. Since this is less than the significance level \(\alpha = 0.05\), it provides sufficient evidence to reject the null hypothesis. This implies that Consultant A's higher experience likely leads to better service ratings.
degrees of freedom
Degrees of freedom are a critical component in statistical calculations, particularly in hypothesis testing. They help determine the shape of the t-distribution, which affects the calculation of the p-value. In two-sample t-tests, the degrees of freedom are calculated through a complex formula that accounts for the variance and sample sizes of the two groups:\[df = \left( \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \right) \]Intuitively, degrees of freedom are related to the number of independent observations reporting useful information. In simpler terms:- More data points afford more degrees of freedom.- More degrees of freedom typically result in a more accurate estimation of the population parameters. In our case, we calculated the degrees of freedom to be approximately 19. After rounding, this value was used to obtain the p-value from the t-distribution table. Accurate degrees of freedom ensure the validity of our statistical conclusion when determining the significance of Consultant A’s higher service ratings.

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Most popular questions from this chapter

Safegate Foods, Inc., is redesigning the checkout lanes in its supermarkets throughout the country and is considering two designs. Tests on customer checkout times conducted at two stores where the two new systems have been installed result in the following summary of the data. \(\begin{array}{cl}\text { System A } & \text { System B } \\ n_{1}=120 & n_{2}=100 \\\ \bar{x}_{1}=4.1 \text { minutes } & \bar{x}_{2}=3.4 \text { minutes } \\\ \sigma_{1}=2.2 \text { minutes } & \sigma_{2}=1.5 \text { minutes }\end{array}\) Test at the .05 level of significance to determine whether the population mean checkout times of the two systems differ. Which system is preferred?

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