Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial consultants and services (2000 Merrill Lynch Client Satisfaction Survey). 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 = .05 and test to see whether the consultant with more experience has the higher population mean service rating.ConsultantA Consultant Bn1=16 n2 =10x1 =6.82 (with bar overx) x2 = 6.25(with bar over x)s1 =0.64 s2 = 0.75a.) State the null and alternative hypothesis.b.) compute the value of the test statistic.c.)What is the p-value?d.) What is your conclusion?

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 AX₁: Rating on the client satisfaction survey given to consultant An₁= 16X[bar]₁= 6.82S₁= 0.64Consultant BX₂: Rating on the client satisfaction survey given to consultant Bn₂= 10X[bar]₂= 6.25S₂= 0.75a. 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 ≤ μ₂H₁: μ₁ > μ₂b. 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.[tex]t_{H_0}= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}[/tex][tex]Sa^2= \frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} = \frac{15*(0.64^2)+9*(0.75^2)}{16+10-2}= 0.47[/tex]Sa= 0.68[tex]t_{H_0}= \frac{6.82-6.25-0}{0.68*\sqrt{\frac{1}{16} +\frac{1}{10} } }= 2.079[/tex]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= 0.0242d. 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!