The Ledd Pipe Company has received a large shipment of pipes, and a quality control inspector wishes to
estimate the average diameter of these pipes. A random sample of 18 pipes produces an average diameter of
2.56 mm with a standard deviation of 0.05 mm.
The average diameter of the pipes must not differ significantly from 2.52 mm. Is there enough evidence at
the 5% level of significance to conclude that the true average diameter differs from this amount? Conduct an
appropriate hypothesis test using (i) the p-value method, (ii) the rejection point method, and (iii) the confidence
interval method.
n = 18
xbar = 2.56 s = 0.05
H0 : mu = 2.52 H1: mu not equal 2.52
test stat = t = (xbar-mu)/(s/sqrt(n)) = 3.39
p-value = 0.003 < 0.05 Reject H0
t-critical = 2.11 < 3.39 Reject H0
CI =(xbar-t-critical*s/sqrt(n), xbar+t-critical*s/sqrt(n))
(2.54, 2.58)
To answer this question, we will conduct a hypothesis test using the p-value method, the rejection point method, and the confidence interval method.
Hypothesis test:
Step 1: State the null and alternative hypotheses.
Null hypothesis (H₀): The true average diameter of the pipes is 2.52 mm.
Alternative hypothesis (H₁): The true average diameter of the pipes is different from 2.52 mm.
Step 2: Determine the test statistic.
We will use the t-test statistic because the population standard deviation is unknown.
Step 3: Set the significance level (α).
The significance level is given as 5%, which corresponds to α = 0.05.
(i) P-value method:
Step 4: Calculate the test statistic.
The test statistic for the difference of means (t-score) can be calculated using the formula:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size.
In this case, x̄ = 2.56 mm, μ = 2.52 mm, s = 0.05 mm, and n = 18.
Substituting these values into the formula gives us:
t = (2.56 - 2.52) / (0.05 / √18)
t ≈ 4.997
Step 5: Calculate the p-value.
We need to find the probability of observing a test statistic more extreme than our calculated t-value. Since the alternative hypothesis is two-sided (true average diameter differs from 2.52 mm), we will calculate the p-value for both tails.
Using a t-distribution table or a statistical software, we find that the two-tailed p-value for t = 4.997 and 16 degrees of freedom is very small, let's say p < 0.001.
Step 6: Compare the p-value to the significance level.
The p-value (p < 0.001) is smaller than the significance level (α = 0.05). Therefore, we reject the null hypothesis.
(ii) Rejection point method:
Step 4: Determine the critical value(s) for the test statistic.
To find the rejection point(s), we compare the calculated test statistic to the critical values of the t-distribution with n-1 degrees of freedom at the desired significance level.
Since this is a two-tailed test, we divide the significance level by 2, resulting in a critical value at α/2 = 0.025.
Using a t-distribution table or a statistical software, we find the critical value for a t-distribution with 16 degrees of freedom and α/2 = 0.025 is approximately ±2.131.
Step 5: Compare the test statistic to the critical value(s).
Our calculated t-value, 4.997, is outside the range of ±2.131. Therefore, we reject the null hypothesis.
(iii) Confidence interval method:
Step 4: Calculate the confidence interval for the mean difference.
We can calculate the confidence interval for the mean difference using the formula:
CI = x̄ ± t * (s / √n)
where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and t is the critical value from the t-distribution corresponding to the desired confidence level.
Assuming a 95% confidence level, the critical value for t with 16 degrees of freedom is approximately ±2.131.
Substituting the values x̄ = 2.56 mm, s = 0.05 mm, n = 18, and t = 2.131 into the formula, we get:
CI = 2.56 ± 2.131 * (0.05 / √18)
CI ≈ 2.56 ± 0.023
Step 5: Interpret the confidence interval.
The confidence interval is (2.537, 2.583) or approximately (2.54 mm, 2.58 mm). Since the hypothesized value of 2.52 mm is not within the confidence interval, we can conclude that there is enough evidence to suggest that the true average diameter differs from this amount.
In summary, using all three methods (p-value method, rejection point method, and confidence interval method), we have found enough evidence to conclude that the true average diameter of the pipes differs from 2.52 mm.