Question 1166031: A sales clerk in the departmental store claims that 65% of the shoppers entering the store leave without making a purchase. A random sample of 60 shoppers showed that 40 of them left without buying anything. Are these sample results consistent with the claim of the sales clerk at 1% level of significance?
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! This problem requires a **one-sample proportion Z-test** to determine if the sample results are consistent with the sales clerk's claim.
## 📝 Hypothesis Test for Population Proportion
### 1. Define Hypotheses
The sales clerk claims that 65% of shoppers leave without purchasing, so the null hypothesis ($H_0$) is that the true population proportion ($p$) is $0.65$. The alternative hypothesis ($H_1$) tests if the sample is *inconsistent* with this claim, suggesting the proportion is different.
* **Null Hypothesis ($H_0$):** $p = 0.65$ (The true proportion of shoppers leaving without a purchase is 65%).
* **Alternative Hypothesis ($H_1$):** $p \neq 0.65$ (The true proportion is different from 65%).
This is a **two-tailed test**.
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### 2. Determine the Level of Significance and Critical Values
* **Level of Significance ($\alpha$):** $1\%$ or $0.01$.
* Since this is a two-tailed test, the significance is split between the two tails ($\alpha/2 = 0.01/2 = 0.005$).
* The **critical Z-values** corresponding to an area of $0.005$ in each tail are found from the standard normal distribution table:
$$Z_{\text{critical}} = \mathbf{\pm 2.58}$$
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### 3. Calculate Test Statistic
We use the sample data to calculate the test statistic ($Z_{\text{calc}}$).
#### A. Sample Proportion ($\hat{p}$)
* Sample size ($n$) = 60
* Number of shoppers leaving without purchase ($x$) = 40
* Sample proportion ($\hat{p}$) is the observed proportion:
$$\hat{p} = \frac{x}{n} = \frac{40}{60} \approx \mathbf{0.6667}$$
#### B. Standard Error ($SE$)
We use the claimed population proportion ($p_0 = 0.65$) to calculate the standard error for the test statistic:
$$SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.65 (1 - 0.65)}{60}}$$
$$SE = \sqrt{\frac{0.65 \times 0.35}{60}} = \sqrt{\frac{0.2275}{60}} \approx \sqrt{0.003792} \approx \mathbf{0.06158}$$
#### C. Calculated Z-Score ($Z_{\text{calc}}$)
$$Z_{\text{calc}} = \frac{\hat{p} - p_0}{SE} = \frac{0.6667 - 0.65}{0.06158}$$
$$Z_{\text{calc}} = \frac{0.0167}{0.06158} \approx \mathbf{0.27}$$
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### 4. Decision and Conclusion
* **Decision Rule:** Reject $H_0$ if $|Z_{\text{calc}}| > |Z_{\text{critical}}|$.
* **Comparison:**
$$|0.27| \text{ is not greater than } |2.58|$$
$0.27 < 2.58$
The calculated test statistic ($Z_{\text{calc}} = 0.27$) falls within the non-rejection region (the area between -2.58 and 2.58).
**Conclusion:**
Since the test statistic is not in the critical region, we **fail to reject the null hypothesis ($H_0$)**.
The sample results ($\hat{p} \approx 66.67\%$) are **consistent with the claim** of the sales clerk ($p = 65\%$) at the $1\%$ level of significance. The small difference observed is attributed to **random sampling variation**.
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