Question

To compare customer satisfaction levels of two competing ice cream companies, 8 customers of Company 1 and 5 customers of Company 2 ware randomly selected and were asked to rate their ice creams on a five-point scale, with 1 being least satisfied and 5 most satisfied. The survey results are summarized in the following table:

Company 1

𝑛1=8

𝑋1 =3.21

 𝑆1= 0.31

Company 2

𝑛2=5

𝑋2=2.22

𝑆2= 0. 21

By assuming both equal variances and unequal variances:

a. Construct an interval estimate at 95°/ confidence interval for difference of means of two ice cream companies.

Answer 1

To test the difference between two means of independent samples with unknown population standard deviations we need to use the t-test.

Confidence interval for the difference of two means will be (unequal variances assumption):

$(\overline{X_1}-\overline{X_2})-t_{\alpha/2}*\sqrt{s_1^2/n_1+s_2^2/n_2}<\mu_1-\mu_2<(\overline{X_1}-\overline{X_2})+t_{\alpha/2}*\sqrt{s_1^2/n_1+s_2^2/n_2}$

d.f. = 5-1 = 4

$t_{\alpha/2}$ (95%, d.f. = 4) = 2.776

$(3.21-2.22)-2.776*\sqrt{0.31^2/8+0.21^2/5}<\mu_1-\mu_2<(3.21-2.22)+2.776*\sqrt{0.31^2/8+0.21^2/5}$

$0.59<\mu_1-\mu_2<1.39$

For equal variances assumption:

$s_p=\sqrt{\frac{(n_1-1)*s_1^2+(n_2-1)*s_2^2}{n_1+n_2-2}}=\sqrt{\frac{(8-1)*0.31^2+(5-1)*0.21^2}{8+5-2}}=0.278$

$(\overline{X_1}-\overline{X_2})-t_{\alpha/2}*s_p*\sqrt{1/n_1+1/n_2}<\mu_1-\mu_2<(\overline{X_1}-\overline{X_2})+t_{\alpha/2}*s_p*\sqrt{1/n_1+1/n_2}$

d.f. = n₁+n₂-2 = 8+5-2 = 11

$t_{\alpha/2}$ (95%, d.f. = 11) = 2.201

$(3.21-2.22)-2.201*0.278*\sqrt{1/8+1/5}<\mu_1-\mu_2<(3.21-2.22)+2.201*0.278*\sqrt{1/8+1/5}$

$0.64<\mu_1-\mu_2<1.34$