The makers of a smartphone have received complaints that the face recognition tool often doesn't work, or takes multiple attempts to finally unlock the phone. They've upgraded to a new version and are claiming the tool has improved. To test the claim, a critic takes a random sample of 80 users of the new version (Group 1) and 75 users of the old version (Group 2). He finds that the face recognition tool works on the first try 70% of the time in the new version and 56% of the time in the old version. Can it be concluded that the new version is performing better? Test at α=0.10.

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.