Question 2
Mean = 300 grams
Standard deviation =10 grams
(ii) The understanding of the empirical rule can be highlighted using the following diagram.
In this case, since µ = 300g and σ = 10g, hence the rejected region would have a probability of 0.025.
The % of data lying between µ- 3σ and µ+3σ in accordance with the empirical rule is 99.7%. However, in order to obtain the answer for the given part, suitable deductions need to be made for the area between µ- σ and µ- 3σ (Lehman & Romano, 2016).
% values lying between µ- 3σ and µ- σ = 13.5% + 2.5% -[(100-99.7)/2]% = 15.85%
Hence, required probability = 0.997-0.1585 = 0.8385
Question 3
As negative skew is present along with potential outlier values speciality on the lower end, hence mean might be distorted to indicate a lower value. Thus, central tendency for the given data would be best represented using median which is not impacted by extreme values on either sides. Further, in relation to deviation, standard deviation may not be reliable and hence IQR or Inter-Quartile Range may be preferred (Shi & Tao, 2015).
(ii) The value of sample mean from the above shown descriptive statistics comes out to be 59.62. Hence, it can be said that values of sample mean and the hypothesized mean are similar.
(iii) The value of t stat
(iv) The p value would be calculated as =1-T.DIST(3.90,299,TRUE) which comes out to be 0.0001.
(v) It can be seen that p value is lower than level of significance (0.02<0.05) and therefore, null hypothesis would be rejected and alternative hypothesis would be accepted (Medhi, 2015).
(vi) It can be concluded that the average satisfaction survey score of clients is not 55.
Question 4
(ii) The key difference as apparent from the above graph is that a higher satisfaction score is witnessed for larger clients in comparison to smaller clients. A key similarity is that majority of the satisfaction survey scores for both big and small clients are concentrated in the middle values with limited concentration in the extremes (Lind, Marchal & Wathen, 2016).
(b) Level of significance = 5%
(i) Hypothesis testing (Harmon, 2015) :
(ii) The two tailed p value comes out to be 0.02.
(iii) It can be seen that p value is lower than level of significance (0.02<0.05) and therefore, null hypothesis would be rejected and alternative hypothesis would be accepted (Lehman & Romano, 2016).
(iv) It may be concluded that the satisfaction level across the big clients and small clients tends to differ which may be indicative of potential differential in providing service to these two groups.
(v) Management needs to link the incentive of project managers with the satisfaction level of clients irrespective of their underlying size. This would ensure that proper attention and customer service would be extended to both big and small clients.
Question 5
Pivot table
Favourable case = 30
Total cases =80
Probability
Favourable case = 20
Total cases =80
Probability
Null hypothesis Variables season and preferences are independent.
Alternative hypothesis Variables season and preferences are dependent.
Chi – square test (Koch, 2016)
(e) During winters, it is apparent that people tend to visit both inland and coastal locations and hence both need to be advertised. This is not the case in summers when coastal locations have a clear edge.
References
Harmon, M. (2015) Hypothesis Testing in Excel – The Excel Statistical Master (7th ed.). Florida: Mark Harmon.
Koch, K.R. (2016) Parameter Estimation and Hypothesis Testing in Linear Models (2nd ed.). London: Springer Science & Business Media.
Lehman, L. E. & Romano, P. J. (2016) Testing Statistical Hypotheses (3rd ed.). Berlin : Springer Science & Business Media.
Lind, A.D., Marchal, G.W. & Wathen, A.S. (2016) Statistical Techniques in Business and Economics (15th ed.). New York : McGraw-Hill/Irwin.
Medhi, J. (2015) Statistical Methods: An Introductory Text (4th ed.). Sydney: New Age International.
Shi, Z. N. & Tao, J. (2015) Statistical Hypothesis Testing: Theory and Methods (6th ed.). London: World Scientific.
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