Question

Company	2007	2008
A	73	77
B	75	72
C	77	76
D	80	84
E	90	88
F	64	69
G	70	75
H	77	79
I	74	82
J	81	85
K	77	72
L	69	76
M	80	86
N	74	77
O	76	73
P	78	79
R	77	76
S	76	78

Will improving customer service result in higher stock prices for the companies providing the better service? “When a company’s satisfaction score has improved over the prior year’s results and is above the national average (currently 75.7), studies show its shares have a good chance of outperforming the broad stock market in the long run” (BusinessWeek, March 2, 2009). The following satisfaction scores of companies for the 4th quarters of 2007 and 2008 were obtained from the American Customer Satisfaction Index. Assume that the scores are based on a poll of 60 customers from each company.

Has the service score improved?

Is the mean score above the national average?

Answer 1

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(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 42 Ha:pi < u2 This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 34. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this left-tailed test is te = -1.691, for a = 0.05 and df = 34. The rejection region for this left-tailed test is R = {t:t< -1.691}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: 71 - X2 (m – 1)of f(n2-1) 3 (1 + +) -1)s +(n2-1) ni +n2-2 76 - 78 = -1.12 V (18-1)5.46652+(18-1)5.24692 18+18-2 ( 18 T 18) (4) Decision about the null hypothesis Since it is observed that t = -1.12 >te = -1.691, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.1353, and since p = 0.1353 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean ui is less than 42, at the 0.05 significance level. Confidence Interval The 95% confidence interval is -5.629 < Mi - M2 < 1.629. Graphically. T-Test: t = -1.12, p = 0.1353 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0