Question 5 options:

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 75% of the time in the new version and 60% of the time in the old version. Can it be concluded that the new version is performing better? Test at α=0.10.

Hypotheses:

H0: *p*1 = *p*2

H1: *p*1 > *p*2

In this scenario, what is the test statistic? Round to four decimal places.

z =_____

To solve this problem, we will perform a hypothesis test for the difference between two proportions. We are given the following information:

- Group 1 (New version): 
  - \( n_1 = 80 \) (sample size)
  - \( p_1 = 0.75 \) (sample proportion)

- Group 2 (Old version): 
  - \( n_2 = 75 \) (sample size)
  - \( p_2 = 0.60 \) (sample proportion)

The null hypothesis (\( H_0 \)) states that the proportions are equal: 
\[
H_0: p_1 = p_2
\]
The alternative hypothesis (\( H_1 \)) states that the proportion for Group 1 is greater than that for Group 2: 
\[
H_1: p_1 > p_2
\]

### Step 1: Compute the pooled proportion
The pooled proportion (\( \hat{p} \)) is calculated as:
\[
\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}
\]
where \( x_1 = p_1 \times n_1 \) and \( x_2 = p_2 \times n_2 \) represent the number of successes in each group.

\[
x_1 = 0.75 \times 80 = 60
\]
\[
x_2 = 0.60 \times 75 = 45
\]
Now calculate the pooled proportion:
\[
\hat{p} = \frac{60 + 45}{80 + 75} = \frac{105}{155} \approx 0.6774
\]

### Step 2: Compute the standard error
The standard error (SE) for the difference in proportions is given by:
\[
SE = \sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}
\]

\[
SE = \sqrt{0.6774 \times (1 - 0.6774) \times \left( \frac{1}{80} + \frac{1}{75} \right)}
\]
\[
SE = \sqrt{0.6774 \times 0.3226 \times \left( \frac{1}{80} + \frac{1}{75} \right)}
\]
\[
SE = \sqrt{0.6774 \times 0.3226 \times 0.02612} \approx \sqrt{0.005674} \approx 0.0753
\]

### Step 3: Compute the test statistic
The test statistic (z) is calculated as:
\[
z = \frac{(p_1 - p_2)}{SE}
\]
\[
z = \frac{(0.75 - 0.60)}{0.0753} = \frac{0.15}{0.0753} \approx 1.992
\]

Thus, the test statistic is approximately \( z = 1.992 \).

Would you like more details or further clarification?

### 5 Related Questions:
1. What is the critical value for a one-tailed test at \( \alpha = 0.10 \)?
2. How do you interpret the results of the z-test in this scenario?
3. How would the test statistic change if the sample size was larger for Group 2?
4. What is the p-value associated with a test statistic of \( z = 1.992 \)?
5. How does the standard error affect the outcome of the hypothesis test?

**Tip:** Always double-check your assumptions and calculations when performing hypothesis tests, especially when calculating pooled proportions and standard errors.

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 75% of the time in the new version and 60% of the time in the old version. Can it be concluded that the new version is performing better? Test at α=0.10.

Hypotheses:

H0: *p*1 = *p*2

H1: *p*1 > *p*2

In this scenario, what is the test statistic? Round to four decimal places.
z =_____

Test whether the upgraded face recognition tool performs better using a Z-test for proportions. Group 1 has a 75% success rate, and Group 2 has a 60% rate. With a sample size of 80 for the new version and 75 for the old version, calculate the test statistic for α=0.10.

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