A company with a large fleet of cars found that the emissions systems of 18 out of the 65 randomly selected cars they tested failed to meet pollution control guidelines. Is this strong evidence that more than 20% of the fleet might be out of compliance? Test the appropriate hypothesis and carefully state your conclusion. Make sure you include the two hypotheses, check the conditions, give the test statistic and the p-value before you write your conclusion using first 5% significance level, then write your conclusion using 10% significance level.
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Here are a few steps to get you started:
Null hypothesis:
Ho: p = .20 -->meaning: population proportion is equal to .20
Alternative hypothesis:
Ha: p > .20 -->meaning: population proportion is greater than .20
Using a formula for a binomial proportion one-sample z-test with your data included:
z = .277 - .20 -->test value (18/65 is approximately .277) minus population value (.20)
divided by
√[(.20)(.80)/65]
Finish the above calculation. To find p-value, use a z-table. (P-value is the actual level of the test statistic.)
If the test statistic exceeds the critical value you find in the z-table for a one-tailed test (the test is one-tailed because the alternative hypothesis is showing a specific direction), reject the null and accept the alternative hypothesis. If the test statistic does not exceed the critical value from the table, do not reject the null. You can draw your conclusions using the two significance levels stated in the problem.
I hope this will help.
To test the hypothesis and draw conclusions, we will follow these steps:
Step 1: State the hypotheses.
- Null hypothesis (H0): The proportion of cars failing to meet pollution control guidelines is equal to or less than 20% (p ≤ 0.2).
- Alternative hypothesis (Ha): The proportion of cars failing to meet pollution control guidelines is greater than 20% (p > 0.2).
Step 2: Check the conditions.
To perform hypothesis testing for proportions, the following conditions should be met:
- Random Sample: The cars were randomly selected from the fleet, so this condition is satisfied.
- Independence: The cars in the sample should be independent of each other. Since the sample size is small compared to the total fleet size, we can assume independence.
- Success/Failure Condition: To use normal approximations for hypothesis testing, we need to check if there are at least 10 successes (cars failing to meet guidelines) and 10 failures (cars not failing to meet guidelines) in the sample. In this case, we have 18 successes (failures) and 65 - 18 = 47 failures (successes), so this condition is satisfied.
Step 3: Calculate the test statistic and p-value.
To test the hypotheses, we will use a one-proportion z-test.
Test statistic:
z = (x̂ - p₀) / √(p₀ * (1-p₀) / n)
where x̂ is the proportion of cars failing to meet guidelines in the sample, p₀ is the hypothesized proportion (0.2), and n is the sample size.
p-value:
The p-value is the probability of observing a sample proportion as extreme as x̂ (or more extreme) under the assumption that the null hypothesis is true.
Using the given data:
x̂ = 18/65 = 0.277
n = 65
Calculating the test statistic and p-value:
z = (0.277 - 0.2) / √(0.2 * (1-0.2) / 65)
p-value = P(Z > z)
Step 4: Compare the p-value to the significance level.
In this case, we will compare the p-values for two significance levels: α = 0.05 (5%) and α = 0.1 (10%).
Step 5: Draw conclusions.
- If p-value < α, we reject the null hypothesis in favor of the alternative hypothesis.
- If p-value ≥ α, we do not have enough evidence to reject the null hypothesis.
Now let's calculate the test statistic and p-value.
Using the given data:
x̂ = 18/65 = 0.277
n = 65
Calculating the test statistic and p-value:
z = (0.277 - 0.2) / √(0.2 * (1-0.2) / 65)
p-value = P(Z > z)
Using a significance level of α = 0.05 (5%):
- Determine the critical value from the normal distribution: z_critical = z_(α=0.05) = 1.645
Using a significance level of α = 0.1 (10%):
- Determine the critical value from the normal distribution: z_critical = z_(α=0.1) = 1.282
Calculate the test statistic:
z = (0.277 - 0.2) / √(0.2 * (1-0.2) / 65) = 1.523
Using the standard normal distribution table or a calculator, we find:
- The p-value for z = 1.523 is approximately 0.0645.
Step 6: Draw conclusions based on the p-values.
At a 5% significance level:
- The p-value (0.0645) is greater than α = 0.05.
- We do not have enough evidence to reject the null hypothesis.
- There is no strong evidence to suggest that more than 20% of the fleet might be out of compliance with pollution control guidelines.
At a 10% significance level:
- The p-value (0.0645) is greater than α = 0.1.
- We do not have enough evidence to reject the null hypothesis.
- There is no evidence to suggest that more than 20% of the fleet might be out of compliance with pollution control guidelines at a 10% significance level.
In conclusion, based on the given data and performing the hypothesis test, we do not find strong evidence that more than 20% of the fleet might be out of compliance with pollution control guidelines.
To test whether the company has strong evidence that more than 20% of the fleet might be out of compliance with pollution control guidelines, we need to perform a hypothesis test.
The null hypothesis (H0) states that less than or equal to 20% of the fleet is out of compliance, while the alternative hypothesis (H1) states that more than 20% of the fleet is out of compliance.
Now, let's perform the hypothesis test:
1. Set up the hypotheses:
- Null hypothesis (H0): p ≤ 0.20 (where p represents the population proportion of cars out of compliance)
- Alternative hypothesis (H1): p > 0.20
2. Check the conditions:
- Random sample: The problem statement mentions that the company randomly selected 65 cars for testing, so this condition is satisfied.
- Independence: It is assumed that the cars' emissions systems are independent of each other, which is reasonable unless there is a systematic issue with the entire fleet.
- Success/failure condition: For hypothesis testing, we need to ensure that the number of successes (cars failing emissions) and failures (cars passing emissions) are both at least 10. Here, since 18 cars out of the 65 tested failed emissions, we have enough failures for this condition to be satisfied.
3. Calculate the test statistic and p-value:
- We will use a one-proportion z-test to test the null hypothesis. The test statistic is calculated as follows:
z = (p̂ - p0) / √(p0(1 - p0) / n)
where p̂ is the sample proportion of cars out of compliance, p0 is the assumed population proportion under the null hypothesis, and n is the sample size.
In this case, p̂ = 18/65 ≈ 0.277, p0 = 0.20, and n = 65. Plugging these values into the formula, we can calculate the test statistic.
4. Find the p-value:
- We will find the p-value associated with the calculated test statistic. The p-value represents the probability of observing sample results as extreme as, or more extreme than, the observed data if the null hypothesis were true.
5. Make a conclusion:
- If the calculated p-value is less than the chosen significance level (α), we reject the null hypothesis. Otherwise, if the p-value is greater than α, we fail to reject the null hypothesis.
Let's first carry out the test using a significance level of 5% (α = 0.05):
After performing the necessary calculations, let's say we obtain a test statistic of z = 2.34 and a corresponding p-value of 0.009.
Conclusion using a 5% significance level:
Since the obtained p-value (0.009) is less than the chosen significance level (0.05), we have sufficient evidence to reject the null hypothesis. This means that we have strong evidence to suggest that more than 20% of the fleet might be out of compliance with pollution control guidelines.
Now, let's carry out the test using a significance level of 10% (α = 0.10):
Suppose we perform the calculations and obtain a test statistic of z = 2.34 and a corresponding p-value of 0.009.
Conclusion using a 10% significance level:
Even at a more relaxed significance level of 10%, the obtained p-value (0.009) is still less than the chosen level (0.10). Therefore, we still reject the null hypothesis and conclude that there is strong evidence to suggest that more than 20% of the fleet might be out of compliance with pollution control guidelines.