Essays on Measurement And Analysis For Business Assignment

Download full paperFile format: .doc, available for editing

The calculated is a t-value =1.697 while critical value is 1.676 and a significance of. 10 which is statistically different at 90% level of significance. Therefore, we accept null hypothesis that sales are comparable to the mean daily sales with 90% confidenceb. Using an F test based on your results and the results from an additional 4 other shopholders (class members – please state the ID's), determine if your day to day weekday sales variation is comparable to the results of the study with 90% confidence. Step 1: There is a difference in Average in daily sales and results of the studyH0: µ1 = µ2H1: the averages are not equal Where µ1 is the Average daily sales, µ2 the results of the studyThe significance level is 90%Calculation F= F= = 1.063Reject H0 if Fcalc < F critDo not reject H0 if Fcalc ≥ F critCritical value at 0.10 or 90% level of significance Reject H0 if Fcalc < F crit (2.205)Do not reject H0 if Fcalc ≥ F crit (2.205)Comparing Since F calc < F crit (1.063 < 2.205), we reject H0.There is sufficient evidence from the sample, which suggest that one of the averages is different from others as suggested by alternative hypothesis.

It means that we accept alternative hypothesis and reject null hypothesis at the 10% level of significance. 5. a. Using a t-test at 80%Conducting significance tests on the independent variable enables us to determine whether the variable is significant in the regression or not. This test involves using a term known as the degree of freedom which is calculated by subtracting the number of variables, including the constant from the number of observations.

The null and alternative hypothesis for the case is as below: H0: (µ1 - µ2) = 0Ha: (µ1< µ2) > 0The significance level is 80%Calculation of the test statisticst= t= = 1.414Step 4: critical value at 0.20 or 80% level of significance Reject H0 if tcalc ≤ t crit = 1.299Step: 5: Comparing Since t calc > tcrit (1.414 1.299), we accept H0The calculated t value of 1.414 and with the confidence interval of 80% the tabulated t value is 1.299. Here i. e., 1.414 > 1.299.

It can be said that the means are not significant and we accept the null hypothesis (). b. Using an F testIs there a significant difference of day to day week day sales variation and the results of the study? Step 1: The null and alternate hypothesis H0 = μ1=μ2. (At least one of the means is different from the other. )Step 2: The Significance level, α Selecting 80% significance level, i.e. α = 0.20Step 3: The value of the Test StatisticF= F= = 1.1707Step 4: Decision (If Fcalc ≤ F crit, reject; otherwise do not reject)Fail to reject null hypothesis, as F-value (1.10707) is less than the F critical of 1.958 at the selected significance level (0.20)Step 5: The result of the hypothesis testAs per our hypothesis test, it has been found that there is no a significant difference of day to day week day sales variation and the results of the study.

6. z- valuez-test divides the differences between the two means obtained in the study by the standard error of differences. An estimate of how much the means should vary on the basis of chance or error” such as seriously showed distribution with the small samples and markedly differing variance.

The use of the t-test often results in probability statement that is extremely accurate, despite the fact that the assumption of homogeneity of variance and normality of underlying distribution are untenable

Download full paperFile format: .doc, available for editing
Contact Us