In: Statistics and Probability
Nine experts rated two brands of coffee in a taste-testing experiment. A rating on a 7-point scale (1equals extremely unpleasing, 7equals extremely pleasing) is given for each of four characteristics: taste, aroma, richness, and acidity. The accompanying data table contains the ratings accumulated over all four characteristics. Complete parts (a) through (d) below.
Expert Brand A Brand B
C.C. 25 26
S.E. 26 26
E.G. 19 22
B.I. 22 24
C.M. 19 21
C.N. 25 26
G.N. 27 26
R.M. 24 26
P.V. 19 21
a. The test statistic is
(Type an integer or a decimal. Round to two decimal places as needed.)
b. What assumption is necessary about the population distribution in order to perform this test?
c. Determine the p-value in (a) and interpret its meaning.
d.Construct a 95% confidence interval estimate of the difference in the mean ratings between the two brands. Recall that mu Subscript Upper DμDequals=mu 1 minus mu 2μ1−μ2, where mu 1μ1 is the mean rating for brand A and mu 2μ2 is the mean rating for brand B.
using excel>addin>phstat<two sample test>pooled variance test
we have
Pooled-Variance t Test for the Difference Between Two Means | ||||
(assumes equal population variances) | ||||
Data | Confidence Interval Estimate | |||
Hypothesized Difference | 0 | for the Difference Between Two Means | ||
Level of Significance | 0.05 | |||
Population 1 Sample | Data | |||
Sample Size | 9 | Confidence Level | 95% | |
Sample Mean | 22.88888889 | |||
Sample Standard Deviation | 3.218867986 | Intermediate Calculations | ||
Population 2 Sample | Degrees of Freedom | 16 | ||
Sample Size | 9 | t Value | 2.1199 | |
Sample Mean | 24.22222222 | Interval Half Width | 2.7870 | |
Sample Standard Deviation | 2.279132389 | |||
Confidence Interval | ||||
Intermediate Calculations | Interval Lower Limit | -4.1203 | ||
Population 1 Sample Degrees of Freedom | 8 | Interval Upper Limit | 1.4537 | |
Population 2 Sample Degrees of Freedom | 8 | |||
Total Degrees of Freedom | 16 | |||
Pooled Variance | 7.7778 | |||
Standard Error | 1.3147 | |||
Difference in Sample Means | -1.3333 | |||
t Test Statistic | -1.0142 | |||
Two-Tail Test | ||||
Lower Critical Value | -2.1199 | |||
Upper Critical Value | 2.1199 | |||
p-Value | 0.3256 | |||
Do not reject the null hypothesis | ||||
a. The test statistic is -1.01
b the necessary assumption about the population distribution in order to perform this test is that the sample should be drawn from the normally distributed population .
c. the p-value is 0.3256 which is large so we can sa that there is no difference in the mean ratings between the two brands.
d.a 95% confidence interval estimate of the difference in the mean ratings between the two brands. is -4.1203 to 1.4537