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 Company 2 ๐1=8 ๐2=5 ๐1 ฬ ฬ ฬ ฬ =3.21 ๐2 ฬ ฬ ฬ ฬ =2.22 ๐1= 0.31 ๐2= 0. 21 By assuming both equal variances and unequal variances: a. Construct an interval estimate at 95ยฐ/ confidence interval for difference of means of two ice cream companies.
We have that
n1 = 8
"\\bar x_1 = 3.21"
s1 = 0.31
n2 = 5
"\\bar x_2 = 2.22"
s2 = 0.21
Assuming both variances are equal:
Since samples are less than 30 in this problem we are dealing withย t-distirbution with n1ย + n2ย โ 2 = 8 + 5 โ 2 = 11 degrees of freedom.
The table t-value for a 95% confidence interval with 25 df is t0.025, 11 = 2.201
The formula forย a 95% confidence interval for the difference of two population means:
where
"s_p=\\sqrt{\\frac{s_1^2(n_1-1)+s_2^2(n_2-1)}{n_1+n_2-2}}""s_p=\\sqrt{\\frac{0.31^2\\cdot7+0.21^2\\cdot4}{8+5-2}}=0.28"
Confidence interval:
"(3.21-2.22)\\pm2.201\\cdot0.28\\sqrt{\\frac{1}{8}+\\frac{1}{5}}=0.99\\pm0.35"
We are 95% confident that the difference in the two population means is between 0.64 and 1.34. Zero is not in this interval so there is a significant difference of means of two ice cream companies.
Constructing a confidence interval for the difference of means when both variances are not equal:
where df is the smaller of n1โ1 and n2โ1
df = min(8-1, 5-1)=4
Therefore the table t-value for a 95% confidence interval with 4 df is t0.025, 4 = 2.776
"(3.21-2.22)\\pm 2.776\\sqrt{\\frac{0.31^2}{8}+\\frac{0.21^2}{5}}=0.99\\pm 0.4"
We are 95% confident that the difference in the two population means is between 0.59 and 1.39. Zero is not in this interval so there is a significant difference of means of two ice cream companies.
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