a) we test the hypotheses;
H0: mean house before is equal to mean change after
H1: mean house before is less than after
Before After d d-"\\bar d" "(d-\\bar d)^2"
128 135 -7 -2.4 5.76
105 110 -5 -0.4 0.16
119 131 -12 -7.4 54.76
140 142 -2 2.6 6.76
98 105 -7 -2.4 5.76
123 130 -7 -2.4 5.76
127 131 -4 0.6 0.36
115 110 5 9.6 92.16
122 125 -3 1.6 2.56
145 149 -4 0.6 0.36
Total -46 174.4
t="\\frac{\\bar d}{\\frac {sd}{\\sqrt n}}"
="\\frac{\\frac{-46}{10}}{\\frac{{\\sqrt\\frac{174.4}{10-1}}}{\\sqrt{10}}}"
t=-3.3045
The t value from the table is 2.8214
since the absolute value of the test statistic is greater than the table value, we reject the null hypothesis in favor of the alternative hypothesis.
b) From the t tables, the corresponding p-value to the test statistic is 0.004. since the p-value is less than 0.01, we reject the null hypothesis.
c) distribution of the differences is assumed to be normal.
d) Yes. this is because the mean of the differences gives a negative value implying the mean before change was lower than the mean after change . Rejection of the null hypothesis also implies that the change is statistically significant.
Comments
Leave a comment