Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test. Ho: p = 0.93 versus H?: p? 0.93 n = 500, x = 450, ? = 0.01 Is npo (1-Po) ? 10? Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a decimal. Do not round.) A. No, because npo (1-Po) = B. Yes, because npo (1-Po) = 32.55 Now find p. ^p = 0.9 (Type an integer or a decimal. Do not round.) Find the test statistic Zo. Zo = -2.63 (Round to two decimal places as needed.) Find the P-value. P-value = 0.009 (Round to three decimal places as needed.) State the conclusion of the hypothesis test. Reject the null hypothesis, because the P-value is less than
The Correct Answer and Explanation is:
To test the hypothesis using the P-value approach, we need to follow these steps:
1. Verify the requirements of the test
The hypothesis test is about the population proportion ppp. We want to test:
- Null hypothesis (H0H_0H0): p=0.93p = 0.93p=0.93
- Alternative hypothesis (HaH_aHa): p≠0.93p \neq 0.93p=0.93
- Sample size (nnn): 500
- Sample successes (xxx): 450
- Significance level (α\alphaα): 0.01
For a z-test for proportions, the requirements are:
- np0(1−p0)≥10n p_0 (1 – p_0) \geq 10np0(1−p0)≥10, where p0p_0p0 is the hypothesized population proportion.
- nnn should be large enough to ensure normality, which is often the case when both np0n p_0np0 and n(1−p0)n (1 – p_0)n(1−p0) are greater than or equal to 10.
Let’s check if np0(1−p0)≥10n p_0 (1 – p_0) \geq 10np0(1−p0)≥10:np0(1−p0)=500×0.93×(1−0.93)=500×0.93×0.07=32.55n p_0 (1 – p_0) = 500 \times 0.93 \times (1 – 0.93) = 500 \times 0.93 \times 0.07 = 32.55np0(1−p0)=500×0.93×(1−0.93)=500×0.93×0.07=32.55
So, the answer is B. Yes, because np0(1−p0)=32.55n p_0 (1 – p_0) = 32.55np0(1−p0)=32.55. This satisfies the requirement for normality.
2. Find the sample proportion
The sample proportion p^\hat{p}p^ is calculated as:p^=xn=450500=0.9\hat{p} = \frac{x}{n} = \frac{450}{500} = 0.9p^=nx=500450=0.9
3. Calculate the test statistic Z0Z_0Z0
The test statistic for a hypothesis test on a population proportion is given by:Z0=p^−p0p0(1−p0)nZ_0 = \frac{\hat{p} – p_0}{\sqrt{\frac{p_0(1 – p_0)}{n}}}Z0=np0(1−p0)p^−p0
Substitute the known values:Z0=0.9−0.930.93×(1−0.93)500Z_0 = \frac{0.9 – 0.93}{\sqrt{\frac{0.93 \times (1 – 0.93)}{500}}}Z0=5000.93×(1−0.93)0.9−0.93Z0=0.9−0.930.93×0.07500Z_0 = \frac{0.9 – 0.93}{\sqrt{\frac{0.93 \times 0.07}{500}}}Z0=5000.93×0.070.9−0.93Z0=−0.030.0651500Z_0 = \frac{-0.03}{\sqrt{\frac{0.0651}{500}}}Z0=5000.0651−0.03Z0=−0.030.0001302=−0.030.01141=−2.63Z_0 = \frac{-0.03}{\sqrt{0.0001302}} = \frac{-0.03}{0.01141} = -2.63Z0=0.0001302−0.03=0.01141−0.03=−2.63
So, the test statistic Z0=−2.63Z_0 = -2.63Z0=−2.63.
4. Find the P-value
The P-value is the probability of observing a value as extreme as or more extreme than the test statistic under the null hypothesis. Since the test is two-tailed (because the alternative hypothesis is p≠0.93p \neq 0.93p=0.93), we calculate the two-tailed P-value as:P=2×P(Z≤−2.63)P = 2 \times P(Z \leq -2.63)P=2×P(Z≤−2.63)
Using a standard normal distribution table or a calculator:P(Z≤−2.63)≈0.0043P(Z \leq -2.63) \approx 0.0043P(Z≤−2.63)≈0.0043
Thus, the two-tailed P-value is:P=2×0.0043=0.0086P = 2 \times 0.0043 = 0.0086P=2×0.0043=0.0086
5. State the conclusion of the hypothesis test
- The P-value 0.00860.00860.0086 is less than the significance level α=0.01\alpha = 0.01α=0.01.
- Since the P-value is less than α\alphaα, we reject the null hypothesis.
Conclusion
Reject the null hypothesis, because the P-value is less than the significance level of 0.01.
