AP Statistics 6.7 Potential Errors When Performing Tests- FRQs - Exam Style Questions

Question

The manager of a large company that sells pet supplies online wants to increase sales by encouraging repeat purchases. The manager believes that if past customers are offered $\$10$ off their next purchase, more than $40$ percent of them will place an order. To investigate the belief, $90$ customers who placed an order in the past year are selected at random. Each of the selected customers is sent an e-mail with a coupon for $\$10$ off the next purchase if the order is placed within $30$ days. Of those who receive the coupon, $38$ place an order.

(a) Is there convincing statistical evidence, at the significance level of $\alpha=0.05$, that the manager’s belief is correct? Complete the appropriate inference procedure to support your answer.

(b) Based on your conclusion from part (a), which of the two errors, Type I or Type II, could have been made? Interpret the consequence of the error in context.

Most-appropriate topic codes (CED):

• TOPIC 6.4: Setting Up a Test for a Population Proportion
• TOPIC 6.6: Concluding a Test for a Population Proportion
• TOPIC 6.7: Potential Errors When Performing Tests

▶️ Answer/Explanation

Detailed solution

(a) Inference Procedure

State:
We will perform a one-sample z-test for a population proportion. Let $p$ be the true proportion of all past customers who would place an order after receiving the coupon. The significance level is $\alpha=0.05$.
The hypotheses are:
$H_0: p = 0.40$
$H_a: p > 0.40$

Plan:
We check the conditions for inference.
1. Random: The data come from a random sample of $90$ customers.
2. Independence (10% condition): It is reasonable to assume the company has more than $10 \times 90 = 900$ past customers.
3. Normality (Large Counts): We check the condition using the null hypothesis value $p_0=0.40$.
$np_0 = 90(0.40) = 36 \ge 10$
$n(1-p_0) = 90(0.60) = 54 \ge 10$
All conditions are met.

Do:
The sample proportion is $\hat{p} = \frac{38}{90} \approx 0.4222$.
The test statistic is:
$z = \frac{\hat{p} – p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{0.4222 – 0.40}{\sqrt{\frac{(0.40)(0.60)}{90}}} \approx \frac{0.0222}{0.0516} \approx 0.430$

The p-value is the area to the right of the test statistic:
p-value = $P(Z > 0.430) \approx 0.3336$

Conclude:
Since the p-value ($0.3336$) is greater than the significance level ($\alpha = 0.05$), we fail to reject the null hypothesis.

There is not convincing statistical evidence to support the manager’s belief that more than $40$ percent of all past customers would place an order after receiving the coupon.

(b) Type of Error

Because we failed to reject the null hypothesis, we could have made a Type II error.

A Type II error in this context would occur if the coupon is actually effective (the true proportion of customers who would place an order is greater than $0.40$), but we failed to find convincing evidence of this. The consequence would be a missed opportunity for the company; they would incorrectly conclude the coupon strategy is not effective and might not implement it, thereby losing potential sales.

AP Statistics 6.7 Potential Errors When Performing Tests- FRQs - Exam Style Questions

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