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An environmental group claims that the 'Gas Guzzler Z' car has a higher rate of ...
Apr 15, 2024
An environmental group claims that the 'Gas Guzzler Z' car has a higher rate of emissions than the average car and should be banned. The average emissions of other cars is 2836 litres per year. In a test of the group's claim, 119 Gas Guzzlers are run for a year. The average emissions in the sample is 3451 . With a significance level of α=0.05\alpha=0.05, is there enough evidence to support the group's claim? You may assume that σ\sigma is 493 . You may find the following R output helpful. You should provide your answer to 2 decimal places. Please show all working. pnorm⁡((xbar−mu)/(\operatorname{pnorm}((x b a r-m u) /( sigma/sqrt(n))) \#\# [1] 1 1−pnorm⁡((xbar−mu)/(1-\operatorname{pnorm}((x b a r-m u) /( sigma/sqrt(n))) \#\# [1] 0 rnorm⁡(mu)[[1]]\operatorname{rnorm}(\mathrm{mu})[[1]] \#\# [1] 1.072989 pnorm (1-(xbar - mu)/sigma) \#\# [1] 0.4022744
Solution by Steps
step 1
Formulate the hypotheses for the test: The null hypothesis H0 H_0 states that the mean emissions of the Gas Guzzler Z car is equal to the average emissions of other cars, μ=2836 \mu = 2836 . The alternative hypothesis Ha H_a states that the mean emissions of the Gas Guzzler Z car is greater than the average emissions of other cars, \mu > 2836
step 2
Calculate the test statistic using the sample mean xˉ \bar{x} , the population mean under the null hypothesis μ \mu , the population standard deviation σ \sigma , and the sample size n n : z=xˉ−μσ/n z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}
step 3
Substitute the given values into the test statistic formula: z=3451−2836493/119 z = \frac{3451 - 2836}{493 / \sqrt{119}}
step 4
Compute the value of the test statistic: z=3451−2836493/119≈8.35 z = \frac{3451 - 2836}{493 / \sqrt{119}} \approx 8.35
step 5
Determine the p-value using the R output for the standard normal distribution: Since the R output for pnorm⁡((xˉ−μ)/(σ/n)) \operatorname{pnorm}((\bar{x} - \mu) / (\sigma/\sqrt{n})) is 1, the p-value for a one-tailed test is 1−1=0 1 - 1 = 0
step 6
Compare the p-value to the significance level α \alpha : Since the p-value ≈0 \approx 0 is less than α=0.05 \alpha = 0.05 , we reject the null hypothesis
Answer
There is enough evidence at the α=0.05 \alpha = 0.05 significance level to support the group's claim that the Gas Guzzler Z car has a higher rate of emissions than the average car.
Key Concept
Hypothesis testing using a one-sample z-test
Explanation
The test statistic calculated from the sample data is significantly higher than the critical value for the given significance level, leading to the rejection of the null hypothesis in favor of the alternative hypothesis.
An environmental group claims that the 'Gas Guzzler Z' car has a higher rate of emissions than the average car and should be banned. The average emissions of other cars is 2836 litres per year. In a test of the group's claim, 119 Gas Guzzlers are run for a year. The average emissions in the sample is 3451 . With a significance level of α=0.05\alpha=0.05, is there enough evidence to support the group's claim? You may assume that σ\sigma is 493 .
Solution by Steps
step 1
State the null hypothesis H0 H_0 and the alternative hypothesis H1 H_1 : H0:μ=2836 H_0: \mu = 2836 , H_1: \mu > 2836
step 2
Calculate the test statistic using the formula: z=xˉ−μσn z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} where xˉ \bar{x} is the sample mean, μ \mu is the population mean under the null hypothesis, σ \sigma is the population standard deviation, and n n is the sample size
step 3
Substitute the given values into the test statistic formula: z=3451−2836493119 z = \frac{3451 - 2836}{\frac{493}{\sqrt{119}}}
step 4
Calculate the value of the test statistic: z=3451−2836493119≈61549310.913≈61545.17≈13.62 z = \frac{3451 - 2836}{\frac{493}{\sqrt{119}}} \approx \frac{615}{\frac{493}{10.913}} \approx \frac{615}{45.17} \approx 13.62
step 5
Determine the p-value using the R output for the standard normal distribution: Since the R output for pnorm⁡ \operatorname{pnorm} of the test statistic is 1, the p-value is 1−1=0 1 - 1 = 0
step 6
Compare the p-value to the significance level α \alpha : Since the p-value ≈0 \approx 0 is less than α=0.05 \alpha = 0.05 , we reject the null hypothesis
Answer
There is enough evidence at the α=0.05 \alpha = 0.05 significance level to support the group's claim that the Gas Guzzler Z car has a higher rate of emissions than the average car.
Key Concept
Hypothesis testing using a one-sample z-test
Explanation
The test statistic calculated is significantly higher than the critical value for a α=0.05 \alpha = 0.05 significance level, leading to the rejection of the null hypothesis and supporting the claim of higher emissions.
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