Answer to Question #194003 in Statistics and Probability for Alice

Question

6 A sample of 40 investment customers serviced by an account manager are found to have had an average of $23,000 in transactions during the past year, with a standard deviation of $8500. A sample of 30 customers serviced by another account manager averaged $28,000 in transactions, with a standard deviation of $11,000. Assuming the population standard deviations are equal, use the 0.05 level of significance in testing whether the population means could be equal for customers serviced by the two account managers. Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for this test? Construct and interpret the 95% confidence interval for the difference between the population means.

Answer 1

40 investment customers are serviced by an account manager, which had an average of $23000 in transactions with a standard deviation of $8500. 30 customers serviced by another manager had an average of $28000 and a standard deviation of $11000. The level of significance is 0.05.

Assuming μ_1 and μ_2 the means of distribution of samples 1 and 2,

A null hypothesis is set up to state that no difference is present.

$H_0: μ_1 = μ_2 \\ H_0: μ_1 − μ_2 = 0$

To state difference, alternate hypothesis is constructed.

$H_1: μ_1 ≠ μ_2 \\ H_1: μ_1 − μ_2 ≠ 0$

Significance level = 0.05

To find the sample test statistic as follows:

$\bar{x_1}=23000 \\ \bar{x_2}=28000 \\ \bar{x_1} -\bar{x_2}=23000 -28000 = -5000$

To find t value as follows:

$s_1 = 8500 \\ s_2=11000 \\ n_1 =40 \\ n_2=30$

From null hypothesis $μ_1 − μ_2 = 0$

$t = \frac{(\bar{x_1} -\bar{x_2})-(\mu_1- \mu_2)}{\sqrt{ \frac{s_1^2}{n_1} + \frac{s^2_2}{n_2}}} \\ = \frac{-5000}{\sqrt{ \frac{8500^2}{40}+ \frac{11000^2}{30} }} = -2.07$

Therefore, p-value is 0.035.

The p-value is less than the significance level, hence H₀ is rejected.

So, the population means cannot be equal for customers serviced by two different managers.

To find 95% confidence interval for $μ_1 − μ_2$ as follows:

For $n_1+n_2, t_{α/2}= 68$ degrees of freedom

α =(1−0.95) = 0.05

From t−table, $t_{α/2} =1.994$

Substituting the values in the following equation:

$CI = (\bar{x_1} -\bar{x_2})±t_{α/2}\sqrt{ \frac{s_1^2}{n_1} + \frac{s^2_2}{n_2}} \\ = -5000±1.994\sqrt{ \frac{8500^2}{40}+ \frac{11000^2}{30} } \\ = -5000±2416.523 \\ =(-7416.523, -2583.477)$

At 95% confidence, the difference will be present between the two boundaries.