Answer:
Step-by-step explanation:
This is a test of 2 population proportions. Let 1 and 2 be the subscript for the new version and the old version. The population proportion of the time that the tool works for the new and old version would be p1 and p2 respectively.
p1 - p2 = difference in the proportion of the times that the tool works for the new and old version
The null hypothesis is
H0 : p1 = p2
p1 - p2 = 0
The alternative hypothesis is
H1 : p1 > p2
p1 - p2 > 0
it is a right tailed test
Sample proportion = x/n
Where
x represents number of success
n represents number of samples
For new version
p1 = 70/100 = 0.7
n1 = 80
x1 = n1p1 = 80 × 0.7 = 56
For old version,
p2 = 56/100 = 0.56
n2 = 75
x2 = n2p2 = 75 × 0.56 = 42
The pooled proportion, pc is
pc = (x1 + x2)/(n1 + n2)
pc = (56 + 42)/(80 + 75) = 0.63
1 - pc = 1 - 0.63 = 0.37
z = (p1 - p2)/√pc(1 - pc)(1/n1 + 1/n2)
z = (0.7 - 0.56)/√(0.63)(0.37)(1/80 + 1/75) = 0.14/0.07759993557
z = 1.8
Since it is a right tailed test, we will look at the area to the right of the z score on the normal distribution table to determine the p value. Therefore,
p = 1 - 0.9641 = 0.04
By using the p value,
Since 0.1 > 0.04, we would reject the null hypothesis. Therefore, at a significance level of 0.1, it can be concluded that the new version is performing better.
