Question #1
a)
We have to use Chi-square test here.
Null hypothesis: There is no association between student’s gender and whether or not they pass BEA674.
Alternative hypothesis: student’s gender and whether or not they pass BEA674 are not independent factors.
Set level of significance at 0.01
Calculate expected frequencies as Column total multiplied by Row Total and divided by Grid total:
Calculate Chi-square statistics:
ℵ2=34-33.10233.10+10-10.90210.90+45-45.90245.90+16-15.10215.10≈0.17
b)
The critical Chi-square value for df=1 and a=0.01 is 6.635. Since 0.17<6.635, we failed to reject the null hypothesis. It seems that the variables are not dependent (at 1% level of significance).
c)
The differences between expected and observed frequencies are quite large. I expect a very large Chi-square value and, hence, the null hypothesis will be rejected. Gender and whether an individual enjoys or not watching action movies are related factors.
Question #2
a)
I use Excel, it is more convenient.
b)
The assumptions of the test are the following:
Observations in groups are independent
My dependent variable (number of pieces of carrot) is approximately normally distributed for each of the four groups
Homogeneity of variances
No significant outliers in the data
Null hypothesis: there is no significant difference in the mean values of pieces of carrot between the four bakeries
Alternative hypothesis: not all mean values of pieces of carrot are approximately equal between the four bakeries.
c)
ANOVA indicates significant difference in pieces of carrot among the four bakeries at 5% level of significance (F=24.425, p<0.001).
Question #3
a)
The dependent variable is the variable that is being predicted. In this case, it is %GDP.
b)
Standard error = SSE/(n-k)=71.691912/32=2.24037225
Df for residual = 33-1 = 32
t-stat for GPD = 11.98343177 (I used the reversed function for Student’s distribution with df=32 and p = 2.27382E-13).
c)
The line of best fit is %GDP = 11.8489 – 0.4845GDP
a)
The value of R-square shows the percentage of variation in response variable that is explained by the model. Approximately 81.78% of variance in %GDP is explained by this regression equation.
b)
This error represents the average distance between observed values and regression line. The average distance in our case is 2.24
c)
Null hypothesis:
H0:β0=β1=0
Alternative hypothesis:Ha:not all beta are equal to 0
The relationship between independent and dependent variable is examined with ANOVA test. It shows that F=143.60, p<0.001. That is why the null hypothesis is rejected.
