Ala rge fleet of cars is maintained with an average of 26 mpg. 50 recent trips displayed a mean of 25.02 mpg and standard deviation of 4.1 mpg. At the 5% level of significance, is their evidence the company has failed to keep their fuel goals? I did the math and got .03380315... but when I put it in calc I got z = 1.69015... It does not compute correctly! I know I do not need p value and .o5 level of sig. = +/- 1.96...but am at a loss how to put this together!
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n (n = # of trips)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
However, from the question asked, don't you want a one-tailed test?
what test are you using? This is very confusing!
It is the Z test (google it). Where is your confusion centered?
whether it is one-tailed or two tailed?
and if one tailed...I do agree with you now that this is the left tail test. However I am not sure how to handle negative value on t. Here is what I did.
I used formula for hypothesis test with mu unknow, t=(xbar-mu)/(sd/sqr n)=(26-25.02)/(4.1/sqr 50)=1.69015..
I am not sure what to do next?
To determine if there is evidence that the company has failed to keep their fuel goals, we can perform a hypothesis test.
Step 1: State the hypothesis
- Null hypothesis (H0): The company has achieved its fuel goal (μ = 26 mpg)
- Alternative hypothesis (Ha): The company has failed to achieve its fuel goal (μ < 26 mpg)
Step 2: Set the significance level
The significance level, also known as alpha (α), represents the probability of making a Type I error. In this case, the significance level is 5%, so α = 0.05.
Step 3: Calculate the test statistic
The test statistic for this hypothesis test is the z-score. The formula to calculate the z-score is:
z = (sample mean - population mean) / (population standard deviation / √n)
In this case:
- Sample mean (x̄) = 25.02 mpg
- Population mean (μ) = 26 mpg
- Population standard deviation (σ) = 4.1 mpg
- Sample size (n) = 50
Plugging in the values, we get:
z = (25.02 - 26) / (4.1 / √50) ≈ -1.69015
Step 4: Determine the critical value
Since the alternative hypothesis is one-sided (μ < 26 mpg), we need to find the critical value for a one-tailed test at a 5% significance level. Since the z-distribution is symmetrical, we can find the critical value by using the Z-table or a calculator. For a one-tailed test with α = 0.05, the critical value is approximately -1.645.
Step 5: Make the decision
If the test statistic (z-score) is less than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the calculated z-score (-1.69015) is more extreme than the critical value (-1.645), so we reject the null hypothesis.
Step 6: Draw the conclusion
Based on our hypothesis test, there is evidence to suggest that the company has failed to keep their fuel goals. The sample provided suggests a mean fuel efficiency lower than the stated goal.
It appears there may have been an error in your calculation. The correct z-score is approximately -1.69015, not 1.69015. Double-check your calculations to ensure accuracy.