Question 1177832: A researcher claims that less than 48% of U.S. cell phone owners use their phone for most of their online browsing. In a random sample of 100 adults, 55% say they use their phone for most of their online browsing. At α=0.05, is there enough evidence to support the researcher's claim?
Parameter: Statistic:
Hypotheses: Significance level:
Check conditions: Test statistic formula:
Work:
Test statistic value:
Critical value:
Conclude:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this hypothesis test step-by-step.
**1. Define Parameter and Statistic**
* **Parameter:** p = the true proportion of U.S. cell phone owners who use their phone for most of their online browsing.
* **Statistic:** p̂ = the sample proportion of adults who use their phone for most of their online browsing.
**2. State Hypotheses and Significance Level**
* **Null Hypothesis (H₀):** p ≥ 0.48 (The proportion is greater than or equal to 48%)
* **Alternative Hypothesis (H₁):** p < 0.48 (The proportion is less than 48% - the researcher's claim)
* **Significance Level (α):** 0.05
**3. Check Conditions**
* **Random Sample:** The problem states a "random sample of 100 adults" was taken.
* **Independence:** We assume that the cell phone browsing habits of one adult are independent of others.
* **Normality (Large Counts):**
* np₀ ≥ 10: 100 * 0.48 = 48 ≥ 10
* n(1 - p₀) ≥ 10: 100 * (1 - 0.48) = 100 * 0.52 = 52 ≥ 10
All conditions are met.
**4. Test Statistic Formula**
* z = (p̂ - p₀) / √[p₀(1 - p₀) / n]
**5. Work and Test Statistic Value**
* p₀ = 0.48
* p̂ = 0.55
* n = 100
* z = (0.55 - 0.48) / √[0.48(1 - 0.48) / 100]
* z = 0.07 / √[0.48(0.52) / 100]
* z = 0.07 / √[0.2496 / 100]
* z = 0.07 / √0.002496
* z = 0.07 / 0.04996
* z ≈ 1.4011
**Test statistic value: z = 1.40**
**6. Critical Value**
* This is a left-tailed test (H₁: p < 0.48).
* For α = 0.05, the critical z-value is -1.645.
**Critical value: -1.645**
**7. Conclude**
* **Decision Rule:** If the test statistic (z) is less than the critical value (-1.645), reject the null hypothesis.
* **Comparison:** Our test statistic (1.40) is greater than the critical value (-1.645).
* **Conclusion:** We fail to reject the null hypothesis. There is not enough evidence to support the researcher's claim that less than 48% of U.S. cell phone owners use their phone for most of their online browsing.
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