Math Problem Statement

Test the hypothesis using the P-value approach: H0: p = 0.51 versus H1: p < 0.51, n = 150, x = 66, α = 0.01. Is np0(1 - p0) ≥ 10? Use technology to find the P-value.

Solution

To solve the hypothesis test problem presented in the image, let's break it down step by step.

1. Hypothesis Setup:

  • Null Hypothesis H0H_0: p=0.51p = 0.51
  • Alternative Hypothesis H1H_1: p<0.51p < 0.51

This is a left-tailed hypothesis test, meaning you are testing if the proportion pp is significantly less than 0.51.

2. Given Values:

  • Sample size n=150n = 150
  • Number of successes x=66x = 66
  • Significance level α=0.01\alpha = 0.01

3. Step 1: Verifying the Requirements

To use a normal approximation, the sample size must satisfy: np0(1p0)10np_0(1 - p_0) \geq 10 where p0=0.51p_0 = 0.51.

Let's calculate: np0=150×0.51=76.5n p_0 = 150 \times 0.51 = 76.5 n(1p0)=150×(10.51)=150×0.49=73.5n (1 - p_0) = 150 \times (1 - 0.51) = 150 \times 0.49 = 73.5 Since both 76.576.5 and 73.573.5 are greater than 10, yes, the condition is satisfied.

4. Step 2: Calculate the Test Statistic

Next, we calculate the sample proportion p^\hat{p}: p^=xn=66150=0.44\hat{p} = \frac{x}{n} = \frac{66}{150} = 0.44

Now, calculate the test statistic zz: z=p^p0p0(1p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} Substitute the values: z=0.440.510.51×0.49150=0.070.001666=0.070.04081.7157z = \frac{0.44 - 0.51}{\sqrt{\frac{0.51 \times 0.49}{150}}} = \frac{-0.07}{\sqrt{0.001666}} = \frac{-0.07}{0.0408} \approx -1.7157

5. Step 3: Find the P-Value

Using technology (z-table or calculator), find the p-value corresponding to z=1.7157z = -1.7157 for a left-tailed test.

Looking up z=1.716z = -1.716 in the z-table gives a p-value of approximately: P(z<1.716)0.043P(z < -1.716) \approx 0.043

6. Step 4: Compare the P-Value to α\alpha

Given:

  • P-value=0.043P\text{-value} = 0.043
  • α=0.01\alpha = 0.01

Since the p-value 0.0430.043 is greater than α=0.01\alpha = 0.01, we fail to reject the null hypothesis.

Conclusion:

The final answer is that the P-value is 0.043, and we do not reject the null hypothesis because the P-value is greater than α\alpha.

Would you like more details on any step?

Here are five related questions to extend your understanding:

  1. How would the hypothesis test change if it were a two-tailed test?
  2. What if the significance level α\alpha was 0.05 instead of 0.01?
  3. How does increasing the sample size affect the result of the hypothesis test?
  4. What assumptions must be met to use the normal approximation in hypothesis testing?
  5. How would you perform this test using a confidence interval approach?

Tip: Understanding the relationship between the p-value and the significance level is crucial to making decisions in hypothesis testing.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
P-Value
Normal Approximation

Formulas

np0(1 - p0) ≥ 10
z = (p̂ - p0) / √(p0(1 - p0) / n)
P(z < z-score)

Theorems

Central Limit Theorem
Normal Distribution Theorem

Suitable Grade Level

College/University Level