Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
It was reported that last year the average price of a gallon gasoline in a city X was $3.15. This year a sample of 50 gas stati...Question
It was reported that last year the average price of a gallon gasoline in a city X was $3.15. This year a sample of 50 gas station had an average price of $3.10 for a gallon. We assume that the population standard deviation of prices is $0.15. We are interested in determining whether this year mean price is less than last year. perform a hypothesis test at the level of significance a=0.05. The assumptions are satisfied:
It is a simple random sample The sample size is so it is more than required number 30.
It is a simple random sample The sample size is so it is more than required number 30.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To perform a hypothesis test to determine if this year's mean price of gasoline in city X is less than last year's average price, we will follow these steps:
### Step 1: Set up the hypotheses
We want to test if this year's mean price is less than last year's mean price. Therefore, we can set up our hypotheses as follows:
- Null hypothesis (\( H_0 \)): \( \mu = 3.15 \) (the mean price this year is equal to last year)
- Alternative hypothesis (\( H_1 \)): \( \mu < 3.15 \) (the mean price this year is less than last year)
### Step 2: Identify the significance level
We are given that the significance level (\( \alpha \)) is 0.05.
### Step 3: Calculate the test statistic
We will use the Z-test for this hypothesis test because we know the population standard deviation. The test statistic for a Z-test can be calculated using the formula:
\[
Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}
\]
Where:
- \( \bar{x} = 3.10 \) (the sample mean)
- \( \mu_0 = 3.15 \) (the population mean under the null hypothesis)
- \( \sigma = 0.15 \) (the population standard deviation)
- \( n = 50 \) (the sample size)
Now we plug in the numbers:
\[
Z = \frac{3.10 - 3.15}{0.15 / \sqrt{50}}
\]
Calculating the denominator:
\[
0.15 / \sqrt{50} \approx 0.15 / 7.071 \approx 0.0212
\]
Now, calculating the Z value:
\[
Z = \frac{-0.05}{0.0212} \approx -2.36
\]
### Step 4: Determine the critical value
For a one-tailed test at \( \alpha = 0.05 \), we can find the critical Z value from the Z-table (standard normal distribution). The critical Z value for \( \alpha = 0.05 \) on the left side is approximately \( -1.645 \).
### Step 5: Make the decision
Now we compare our calculated Z value with the critical value:
- If \( Z < -1.645 \), we reject the null hypothesis.
- If \( Z \geq -1.645 \), we fail to reject the null hypothesis.
In our case:
\[
Z \approx -2.36 < -1.645
\]
### Conclusion
Since the calculated Z value is less than the critical Z value, we reject the null hypothesis.
### Final Statement
At the 0.05 significance level, there is sufficient evidence to conclude that the mean price of gasoline this year is less than the mean price of gasoline last year in city X.
### Step 1: Set up the hypotheses
We want to test if this year's mean price is less than last year's mean price. Therefore, we can set up our hypotheses as follows:
- Null hypothesis (\( H_0 \)): \( \mu = 3.15 \) (the mean price this year is equal to last year)
- Alternative hypothesis (\( H_1 \)): \( \mu < 3.15 \) (the mean price this year is less than last year)
### Step 2: Identify the significance level
We are given that the significance level (\( \alpha \)) is 0.05.
### Step 3: Calculate the test statistic
We will use the Z-test for this hypothesis test because we know the population standard deviation. The test statistic for a Z-test can be calculated using the formula:
\[
Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}
\]
Where:
- \( \bar{x} = 3.10 \) (the sample mean)
- \( \mu_0 = 3.15 \) (the population mean under the null hypothesis)
- \( \sigma = 0.15 \) (the population standard deviation)
- \( n = 50 \) (the sample size)
Now we plug in the numbers:
\[
Z = \frac{3.10 - 3.15}{0.15 / \sqrt{50}}
\]
Calculating the denominator:
\[
0.15 / \sqrt{50} \approx 0.15 / 7.071 \approx 0.0212
\]
Now, calculating the Z value:
\[
Z = \frac{-0.05}{0.0212} \approx -2.36
\]
### Step 4: Determine the critical value
For a one-tailed test at \( \alpha = 0.05 \), we can find the critical Z value from the Z-table (standard normal distribution). The critical Z value for \( \alpha = 0.05 \) on the left side is approximately \( -1.645 \).
### Step 5: Make the decision
Now we compare our calculated Z value with the critical value:
- If \( Z < -1.645 \), we reject the null hypothesis.
- If \( Z \geq -1.645 \), we fail to reject the null hypothesis.
In our case:
\[
Z \approx -2.36 < -1.645
\]
### Conclusion
Since the calculated Z value is less than the critical Z value, we reject the null hypothesis.
### Final Statement
At the 0.05 significance level, there is sufficient evidence to conclude that the mean price of gasoline this year is less than the mean price of gasoline last year in city X.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.