Question: The speed with which consumers decide to purchase a product was investigated in the Journal of Consumer Research (August 2011). The researchers theorized that consumers with last names that begin with letters later in the alphabet will tend to acquire items faster than those whose last names begin with letters earlier in the alphabet—called the last name effect. MBA students were offered free tickets to an event for which there was a limitedsupply of tickets. The first letter of the last name of those who responded to an email offer in time to receive the tickets was noted as well as the response time (measured in minutes). The researchers compared the response times for two groups of MBA students: (1) those with last names beginning with one of the first nine letters of the alphabet and (2) those with last names beginning with one of the last nine letters of the alphabet. Summary statistics for the two groups are provided in the table.

	First 9 Letters: A–I	Last 9 Letters: R–Z
Sample size	25	25
Mean response time (minutes)	25.08	19.38
Standard deviation (minutes)	10.41	7.12

Source: Based on K. A. Carlson and J. M. Conrad, “The Last Name Effect: How Last Name Influences Acquisition Timing,” Journal of Consumer Research, Vol. 38, No. 2, August 2011.

a. Construct a 95% confidence interval for the difference between the true mean response times for MBA students in the two groups.

b. Based on the interval, part a, which group has the shorter mean response time? Does this result support the researchers’ last name effect theory? Explain.

Step by step solution

01

(a) Find confidence interval.

$$\begin{gathered} n_1=25,n_2=25, \\ {\overline{x}}_1=25.08,{\overline{x}}_2=19.38 \\ {\sigma }_1=10.41,{\sigma }_2=7.12 \end{gathered}$$

The degree of freedom will be

$$\begin{gathered} =n_1+n_2-2 \\ =25+25-2 \\ =48 \end{gathered}$$

From the t-distribution table, the critical value at 0.05 thelevel of significance for 48 degrees of freedom is 2.011.

The pooled standard deviation is

$$\begin{gathered} s_p=\sqrt{\frac{\left(n_1-1\right){{\sigma }_1}^2+\left(n_2-1\right){{\sigma }_2}^2}{n_1+n_2-2}} \\ =\sqrt{\frac{\left(25-1\right){\left(10.41\right)}^2+\left(25-1\right){\left(7.12\right)}^2}{25+25-2}} \\ =\sqrt{\frac{3817.5}{48}} \\ =8.92 \end{gathered}$$

The 95% confidence interval for the difference in means

$$\begin{gathered} =\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \\ =\left(25.08-19.38\right)\pm 2.011\times 8.92\sqrt{\frac{1}{25}+\frac{1}{25}} \\ =5.7\pm 5.07 \end{gathered}$$

Thus, the confidence interval for means difference is$0.63\text{to 10.77}$

02

(b) Give a conclusion.

The 95%confidence interval contains only positive numbers and does not contain zero, which shows that there is a significant difference between the two groups.

This result supports the researcher's "last name effect" theory that the mean response time for the students whose last names begin with the letters R - Z is shorter than the mean response time for the students whose last names begin with the letters A - I.

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