Hinton Press hypothesizes that the average life of its largest web press is 14,500 hours. They know
that the standard deviation of press life is 2,100 hours. From a sample of 25 presses, the company
finds a sample mean of 13000 hours. At a 0.01 significance level, should the company conclude that
the average life of the presses is less than the hypothesised 14500 hours?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
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.
Well, if I were a web press, I'd say that's a pretty tough act to follow. But let's see if we can figure this out humorously.
First, let's state our hypotheses:
H0: The average life of the presses is 14500 hours.
H1: The average life of the presses is less than 14500 hours.
Next, we need to calculate the test statistic. In this case, we'll use a one-sample t-test because the sample size is less than 30 and we don't know the population standard deviation.
Given that the sample mean is 13000 hours, the population mean is hypothesized to be 14500 hours, and the sample standard deviation is 2100 hours, we can calculate the t-score like this:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
Plugging in the values, we get:
t = (13000 - 14500) / (2100 / sqrt(25))
t = -1500 / (2100 / 5)
t = -1500 / 420
t ≈ -3.57
Now, we need to find the critical t-value for a one-tailed test with a significance level of 0.01 and degrees of freedom (df = sample size - 1 = 25 - 1 = 24).
Using a t-table or calculator, we find the critical t-value to be approximately -2.797.
Since our calculated t-value (-3.57) is lower than the critical t-value (-2.797), we can conclude, with 99% confidence, that the average life of the presses is statistically significantly less than the hypothesized 14500 hours.
So, it seems Hinton Press might need to press pause and reconsider their expectations. But hey, that's just statistically speaking!
To determine whether the company should conclude that the average life of the presses is less than the hypothesized 14,500 hours, we can perform a hypothesis test using the given information.
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H₀): The average life of the presses is equal to or greater than 14,500 hours. (μ ≥ 14,500)
- Alternative hypothesis (H₁): The average life of the presses is less than 14,500 hours. (μ < 14,500)
Step 2: Determine the significance level:
The given significance level is 0.01, which corresponds to a confidence level of 1 - 0.01 = 0.99.
Step 3: Compute the test statistic:
The test statistic for this hypothesis test is the z-score, which can be calculated using the formula:
z = (x̄ - μ) / (σ / √n)
where:
x̄ = sample mean
μ = hypothesized mean
σ = standard deviation
n = sample size
Given:
x̄ = 13,000 hours
μ = 14,500 hours
σ = 2,100 hours
n = 25
Substituting these values into the formula:
z = (13,000 - 14,500) / (2,100 / √25)
z = -1500 / (2100 / 5)
z = -1500 / 420
z ≈ -3.5714
Step 4: Determine the rejection region:
Since the alternative hypothesis is that the average life of the presses is less than 14,500 hours, we are conducting a left-tailed test. At a significance level of 0.01, the critical z-value is approximately -2.33.
Step 5: Make a decision:
The test statistic, -3.5714, falls in the rejection region (z < -2.33), indicating that we reject the null hypothesis.
Step 6: State the conclusion:
Based on the sample data and the hypothesis test, there is sufficient evidence to conclude, at a significance level of 0.01, that the average life of the presses is less than the hypothesized 14,500 hours.
To answer this question, we can perform a one-sample t-test. The null hypothesis, denoted as H0, is that the average life of the presses is equal to 14,500 hours. The alternative hypothesis, denoted as Ha, is that the average life is less than 14,500 hours.
To perform the t-test, we need to calculate the test statistic and compare it to the critical value. The test statistic is calculated using the formula:
t = (sample mean - hypothesized mean) / (standard deviation / sqrt(sample size))
In this case, the sample mean is 13,000 hours, the hypothesized mean is 14,500 hours, the standard deviation is 2,100 hours, and the sample size is 25.
Plugging in the values, we get:
t = (13000 - 14500) / (2100 / sqrt(25)) = -500 / (2100 / 5) ≈ -500 / 420 ≈ -1.19
The next step is to determine the critical value. Since we have a significance level of 0.01 (or 1%), we can refer to a t-distribution table or use statistical software to find the critical value. At a 0.01 significance level with a sample size of 25, the critical value is approximately -2.485.
Finally, we can compare the test statistic to the critical value. If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, |-1.19| = 1.19 < 2.485, so the test statistic does not exceed the critical value. Therefore, we fail to reject the null hypothesis.
In conclusion, based on the sample data and a significance level of 0.01, there is not enough evidence to conclude that the average life of the presses is less than 14,500 hours.