Service Marketing
- Submitted by: kapo0630
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- Date Submitted: 06/05/2013 11:53 PM
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a. Let µ be the mean number of hours of online game playing
H0: µ > 7.7
H1: µ ≤ 7.7
The level of significanceα is given to be 0.05.
The sample size is given to be 200
Since the population is approximately normally distributed and the population standard deviation is known, a Z test can be used and the corresponding test statics is:
The critical value of Z is the value which divides the rejection and non-rejection regions. In this case, the critical value is the Z value to the left of which lies 0.05 of the area under the standard normal curve. From table, - Zα = - Z0.05 = - 1.645
The null hypothesis will be rejected if the test statistic is relatively small, i.e., if the test statistic is less than – 1.645, H0 will be rejected.
Relevant data was given in the question. The value of the test statistic from the data is
8.25 – 7.70
Z = ---------------- = 2.222
3.5 / √200
Z = 2.222 is in the non-rejection region, since 2.222 > - 1.645.
We cannot reject the null hypothesis.
b. Dependent variable is ‘preference for jackets’.
Independent variables are ‘Comfort’, ‘Style’, and ‘Durability’.
The R-square is high which is 0.741. It means that 74.1% of the variation in ‘preference for jackets’ can be explained from the independent variables.
The results in the ANOVA table also indicate that the overall model is significant because the p-value (0.004) is < 0.05 level of significance. This implies that there is evidence to believe that the model is valid.
Multiple regression equation is as follows:
Preference = - 0.539 + 0.258 Comfort + 0.504 Style + 0.336 Durability
The partial regression coefficients indicate that a unit change in ‘Comfort’ will increase ‘preference for jackets’ by 0.258 units while others are held constant. A unit change in ‘Style’ will increase ‘preference for jackets’ by 0.504 units while others are held constant. A unit change in ‘Durability’ will also increase ‘preference for jackets’ by 0.336 units...
H0: µ > 7.7
H1: µ ≤ 7.7
The level of significanceα is given to be 0.05.
The sample size is given to be 200
Since the population is approximately normally distributed and the population standard deviation is known, a Z test can be used and the corresponding test statics is:
The critical value of Z is the value which divides the rejection and non-rejection regions. In this case, the critical value is the Z value to the left of which lies 0.05 of the area under the standard normal curve. From table, - Zα = - Z0.05 = - 1.645
The null hypothesis will be rejected if the test statistic is relatively small, i.e., if the test statistic is less than – 1.645, H0 will be rejected.
Relevant data was given in the question. The value of the test statistic from the data is
8.25 – 7.70
Z = ---------------- = 2.222
3.5 / √200
Z = 2.222 is in the non-rejection region, since 2.222 > - 1.645.
We cannot reject the null hypothesis.
b. Dependent variable is ‘preference for jackets’.
Independent variables are ‘Comfort’, ‘Style’, and ‘Durability’.
The R-square is high which is 0.741. It means that 74.1% of the variation in ‘preference for jackets’ can be explained from the independent variables.
The results in the ANOVA table also indicate that the overall model is significant because the p-value (0.004) is < 0.05 level of significance. This implies that there is evidence to believe that the model is valid.
Multiple regression equation is as follows:
Preference = - 0.539 + 0.258 Comfort + 0.504 Style + 0.336 Durability
The partial regression coefficients indicate that a unit change in ‘Comfort’ will increase ‘preference for jackets’ by 0.258 units while others are held constant. A unit change in ‘Style’ will increase ‘preference for jackets’ by 0.504 units while others are held constant. A unit change in ‘Durability’ will also increase ‘preference for jackets’ by 0.336 units...