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Business leaders around the world are becoming aware of the huge potential of d...
Aug 7, 2024
Solution by Steps
step 1
We will conduct a chi-square test for independence to determine if there is a significant difference in the proportion of organizations that have embarked on digital transformation based on industry sector. The null hypothesis H0 H_0 states that there is no difference in proportions across sectors, while the alternative hypothesis Ha H_a states that there is a difference
step 2
We will calculate the expected frequencies for each cell in the table using the formula: Eij=(row totali)(column totalj)grand total E_{ij} = \frac{(row\ total_i)(column\ total_j)}{grand\ total} where Eij E_{ij} is the expected frequency for the i i -th row and j j -th column
step 3
The observed frequencies Oij O_{ij} and expected frequencies Eij E_{ij} will be used to calculate the chi-square statistic: χ2=(OijEij)2Eij \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} We will then compare this statistic to the critical value from the chi-square distribution with the appropriate degrees of freedom
step 4
The degrees of freedom for this test can be calculated as: df=(r1)(c1) df = (r - 1)(c - 1) where r r is the number of rows and c c is the number of columns in the contingency table. In this case, df=(51)(21)=4 df = (5 - 1)(2 - 1) = 4
step 5
Using a chi-square table or calculator, we will find the critical value for df=4 df = 4 at the 0.05 significance level, which is approximately 9.488. If our calculated χ2 \chi^2 exceeds this value, we will reject the null hypothesis
Answer
If the calculated chi-square statistic is greater than 9.488, we conclude that there is evidence of a difference in the proportion of organizations that have embarked on digital transformation based on industry sector.
Key Concept
Chi-square test for independence is used to determine if there is a significant association between two categorical variables.
Explanation
The chi-square test compares observed and expected frequencies to assess whether the distribution of responses differs significantly across categories.
Solution by Steps
step 1
To analyze the data from Table AMS 10.1, we will conduct an independent samples t-test to compare the update times of the two email interfaces. The null hypothesis H0 H_0 states that there is no difference in the mean update times between the two interfaces, while the alternative hypothesis Ha H_a states that there is a difference
step 2
We will calculate the means and standard deviations for both email interfaces. Let X1ˉ \bar{X_1} and X2ˉ \bar{X_2} be the means for Email Interface 1 and Email Interface 2, respectively, and s1 s_1 and s2 s_2 be their standard deviations
step 3
The t-statistic can be calculated using the formula: t=X1ˉX2ˉs12n1+s22n2 t = \frac{\bar{X_1} - \bar{X_2}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} where n1 n_1 and n2 n_2 are the sample sizes for each group (both are 15)
step 4
We will compare the calculated t-statistic to the critical t-value from the t-distribution table at a significance level of 0.05 with n1+n22 n_1 + n_2 - 2 degrees of freedom. If the absolute value of the t-statistic is greater than the critical value, we reject the null hypothesis
step 5
For the second scenario, where each subscriber's update time is measured for both systems, we will conduct a paired samples t-test. The null hypothesis remains the same, but we will use the differences in update times for each subscriber to calculate the t-statistic
step 6
The t-statistic for the paired samples can be calculated using the formula: t=DˉsD/n t = \frac{\bar{D}}{s_D/\sqrt{n}} where Dˉ \bar{D} is the mean of the differences, sD s_D is the standard deviation of the differences, and n n is the number of pairs (15)
Answer
The analysis will determine if there is a significant difference in update times between the two email interfaces based on the chosen statistical tests.
Key Concept
Independent samples t-test and paired samples t-test are used to compare means between two groups.
Explanation
The independent samples t-test is appropriate for comparing two different groups, while the paired samples t-test is used when the same subjects are measured under two conditions.
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