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 𝑛1=8 𝑛2=5 𝑋1 ̅̅̅̅=3.21 𝑋2 ̅̅̅̅=2.22 𝑆1= 0.31 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.
We have that
n1 = 8
s1 = 0.31
n2 = 5
s2 = 0.21
Assuming both variances are equal:
Since samples are less than 30 in this problem we are dealing with t-distirbution with n1 + n2 – 2 = 8 + 5 – 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 for a 95% confidence interval for the difference of two population means:
where
Confidence interval:
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 n1–1 and n2–1
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
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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