QUESTION I
H0: μ=4.7
H0: μ ≠ 4.7
It is a two tailed test. This is because the test intends to measure whether the average satisfaction score is below the lower tolerance limit (μ<4.7) or the average satisfaction score above the upper tolerance limit (μ>4.7)
Calculated t-score = (x-μ)/sẋ
Where; μ is the target market average score which is 4.5
X is the sample average score which is 4.7
sx is the standard error for the sample which is given by standard deviation of the tested
sample which is 0.25 divided by the square root of the sample size.
sx = 0.25/√20 =0.056
Therefore;
Calculated t-score = (4.7-4.5)/ 0.056 = 3.58
At 95% level of confidence for a two tailed test, the t-critical from the t-table is 2.26
The null hypothesis is rejected. At 95% level of confidence, t-calculated is greater than t-critical. That is 3.58˃2.26.
The most likely satisfaction score is 4.7
Rejecting the null hypothesis implies there is difference between the consumers average satisfaction score from the sample and the target market satisfaction score is statistically significant. The most likely customer satisfaction score of the new product in the whole target market.
QUESTION II
Co-efficient of variation is the ratio of standard deviation of a given data set to the mean of the data. Co-efficient of variation is an important statistical tool for analyzing data about the new product satisfaction scores since it can be used to compare data sets with widely different means and it is an absolute measure.
References
Gibilisco, S. (2004). Statistics demystified (illustrated ed.). New York: McGraw-Hill Professional.
Healey, J. F. (2011). Statistics: A Tool for Social Research. London: Cengage Learning.
Johnson, R. A., & Bhattacharyya, G. K. (2009). Statistics: Principles and Methods (6 ed.). New York: John Wiley and Sons.
