Question 1167903: 2. A Corkill Machine is set ti fill a small bottle with 9.0 grams of medicine. It is claimed that the mean weight is less 9.0 grams. The hypothesis is to be tested at the 0.01 level. A sample revealed these weights (in grams): 9.2, 8.7, 8.9, 8.6, 8.8, 8.5, 8.7, and 9.0.
a. State the null and alternate hypotheses.
b. How many degrees of freedom are there?
c. Give the decision rule.
d. Compute t-value and arrive at a decision.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! This problem requires a **one-sample t-test** because we are comparing a sample mean to a known population mean (or hypothesized mean) when the population standard deviation is unknown and the sample size is small ($n < 30$).
**Given Data:**
* Hypothesized population mean ($\mu_0$): 9.0 grams
* Sample weights: $X = \{9.2, 8.7, 8.9, 8.6, 8.8, 8.5, 8.7, 9.0\}$
* Sample size ($n$): 8
* Significance level ($\alpha$): 0.01
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**a. State the null and alternate hypotheses.**
* **Null Hypothesis ($H_0$):** The mean weight of medicine filled by the Corkill Machine is equal to or greater than 9.0 grams.
$H_0: \mu \ge 9.0$
* **Alternate Hypothesis ($H_1$):** The mean weight of medicine filled by the Corkill Machine is less than 9.0 grams.
$H_1: \mu < 9.0$
This is a **one-tailed (left-tailed) test**.
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**b. How many degrees of freedom are there?**
For a one-sample t-test, the degrees of freedom ($df$) are calculated as $n - 1$, where $n$ is the sample size.
* $df = 8 - 1 = 7$
There are **7 degrees of freedom**.
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**c. Give the decision rule.**
The decision rule is based on comparing the calculated t-value to a critical t-value from the t-distribution table.
* **Significance Level ($\alpha$):** 0.01
* **Degrees of Freedom ($df$):** 7
* **Type of Test:** One-tailed (left-tailed)
From the t-distribution table, the critical t-value for $\alpha = 0.01$ and $df = 7$ for a one-tailed test is approximately **-2.998**.
**Decision Rule:** Reject the null hypothesis ($H_0$) if the calculated t-value is less than -2.998. Otherwise, fail to reject $H_0$.
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**d. Compute t-value and arrive at a decision.**
**1. Calculate Sample Statistics:**
* **Sample Mean ($\bar{x}$):**
$\sum X = 9.2 + 8.7 + 8.9 + 8.6 + 8.8 + 8.5 + 8.7 + 9.0 = 70.4$
$\bar{x} = \frac{\sum X}{n} = \frac{70.4}{8} = 8.8$
* **Sample Standard Deviation ($s$):**
First, calculate $\sum (X - \bar{x})^2$:
* $(9.2 - 8.8)^2 = (0.4)^2 = 0.16$
* $(8.7 - 8.8)^2 = (-0.1)^2 = 0.01$
* $(8.9 - 8.8)^2 = (0.1)^2 = 0.01$
* $(8.6 - 8.8)^2 = (-0.2)^2 = 0.04$
* $(8.8 - 8.8)^2 = (0.0)^2 = 0.00$
* $(8.5 - 8.8)^2 = (-0.3)^2 = 0.09$
* $(8.7 - 8.8)^2 = (-0.1)^2 = 0.01$
* $(9.0 - 8.8)^2 = (0.2)^2 = 0.04$
$\sum (X - \bar{x})^2 = 0.16 + 0.01 + 0.01 + 0.04 + 0.00 + 0.09 + 0.01 + 0.04 = 0.36$
$s = \sqrt{\frac{\sum (X - \bar{x})^2}{n-1}} = \sqrt{\frac{0.36}{8-1}} = \sqrt{\frac{0.36}{7}} \approx \sqrt{0.051428} \approx 0.22678$
**2. Compute the t-value:**
The formula for the t-statistic is:
$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$
$t = \frac{8.8 - 9.0}{0.22678 / \sqrt{8}}$
$t = \frac{-0.2}{0.22678 / 2.8284}$
$t = \frac{-0.2}{0.08018}$
$t \approx -2.494$
**3. Arrive at a Decision:**
* Calculated t-value: $-2.494$
* Critical t-value: $-2.998$
Since the calculated t-value ($-2.494$) is **greater than** the critical t-value ($-2.998$), it does not fall into the rejection region.
**Decision:** Fail to reject the null hypothesis ($H_0$).
**Conclusion:**
At the 0.01 significance level, there is **not enough statistical evidence** to support the claim that the mean weight of medicine filled by the Corkill Machine is less than 9.0 grams.
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