SOLUTION: A sales man in a departmental store claims that 55% of the shoppers entering the store leaves without purchasing anything, a random sample of 50 shoppers showed that 40 of them lef
Algebra.Com
Question 1184731: A sales man in a departmental store claims that 55% of the shoppers entering the store leaves without purchasing anything, a random sample of 50 shoppers showed that 40 of them left without buying anything can we accept clerks claim that at 5% level of significance.
Calculate test value and critical value
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to perform a hypothesis test for the salesperson's claim:
**1. State the Hypotheses:**
* **Null Hypothesis (H0):** The proportion of shoppers leaving without purchasing is 55% (p = 0.55).
* **Alternative Hypothesis (H1):** The proportion of shoppers leaving without purchasing is *not* 55% (p ≠ 0.55). This is a two-tailed test.
**2. Determine the Level of Significance:**
* α = 0.05 (given)
**3. Calculate the Sample Proportion:**
* Sample proportion (p̂) = (Number of shoppers who left without buying) / (Total number of shoppers in the sample) = 40 / 50 = 0.80
**4. Calculate the Test Statistic (z-score):**
* z = (p̂ - p) / sqrt[(p * (1 - p)) / n]
* z = (0.80 - 0.55) / sqrt[(0.55 * (1 - 0.55)) / 50]
* z = 0.25 / sqrt(0.00495)
* z = 0.25 / 0.07036
* z ≈ 3.55
**5. Determine the Critical Value(s):**
* Since this is a two-tailed test with α = 0.05, we need to find the z-scores that correspond to the area in the tails being 0.025 each (0.05 / 2).
* Using a z-table or calculator, the critical values are approximately ±1.96.
**6. Compare the Test Statistic to the Critical Values:**
* Our test statistic (z ≈ 3.55) falls outside the critical region (beyond ±1.96).
**7. Make a Decision:**
* Because the test statistic falls in the rejection region, we reject the null hypothesis.
**8. Conclusion:**
There is sufficient evidence at the 5% level of significance to conclude that the salesperson's claim is incorrect. The sample data suggests that the proportion of shoppers leaving without purchasing is significantly different from 55%. In fact, the sample suggests it is much higher.
RELATED QUESTIONS
A sales clerk in a departmental store claims that 60 % of the shoppers entering the store (answered by stanbon)
A sales clerk in the departmental store claims that 65% of the shoppers entering the... (answered by Boreal)
A sales clerk in a departmental store claims that 60%of the shoppers entering the store... (answered by Theo)
A sales clerk in a departmental store claims that 45% of the shoppers entering the store... (answered by Boreal)
Store gives a discount to 4 of every 10 shoppers. What percent of the shoppers would... (answered by nerdybill)
A large department store has on average 448 shoppers in an hour, with a standard... (answered by Boreal)
SUPPOSE THAT DURING ANY HOUR IN A LARGE DEPARTMENT STORE, THE AVERAGE NUMBER OF SHOPPERS... (answered by stanbon)
A quarter of the shoppers in a store have a loyalty card. Calculate the probability that... (answered by ewatrrr)
The average sales of a product for a particular weel, excluding sunday, were reported by... (answered by stanbon)