Cough Syrup Is Supposed To Contain 6 Ounces Of Medicine
Cough Syrup Is Supposed To Contain 6 Ounces Of Medicine Per Bottle Ho
Cough syrup is supposed to contain 6 ounces of medicine per bottle. However, since the filling machine is not always precise, there can be variation from bottle to bottle. The amounts in the bottles are normally distributed with a standard deviation (σ) of 0.3 ounces. A quality assurance inspector measures 10 bottles and finds the following amounts (in ounces): 5.95, 6.10, 5.98, 6.01, 6.25, 5.85, 5.91, 6.05, 5.88, 5.91.
The question is whether these results provide sufficient evidence to conclude that the bottles are not filled adequately at the labeled amount of 6 ounces per bottle.
Paper For Above instruction
Introduction
Ensuring product consistency and quality in manufacturing processes is crucial for maintaining consumer trust and complying with safety standards. In this context, a quality assurance inspector tests whether bottles of cough syrup contain the advertised 6 ounces of medicine. Variations in filling machines can lead to underfilled or overfilled bottles, potentially affecting product integrity and customer satisfaction. This paper examines whether the observed measurements provide evidence that the actual mean fill is less than the labeled amount, using hypothesis testing methods.
Hypotheses Formulation
The first step in the statistical analysis involves defining the null hypothesis (H₀) and an alternative hypothesis (H₁). We formulate:
- Null hypothesis (H₀): The true mean amount of cough syrup per bottle equals 6 ounces, i.e., μ = 6.
- Alternative hypothesis (H₁): The true mean amount of cough syrup per bottle is less than 6 ounces, i.e., μ
This is a one-tailed test aimed at determining whether the filling process results in underfilled bottles. The significance level (α) is conventionally set at 0.05, meaning we are willing to accept a 5% chance of wrongly rejecting the null hypothesis.
Sample Data and Descriptive Statistics
The inspector measured 10 bottles with the following amounts:
5.95, 6.10, 5.98, 6.01, 6.25, 5.85, 5.91, 6.05, 5.88, 5.91.
Calculating the sample mean (\(\bar{x}\)):
\[
\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{5.95 + 6.10 + 5.98 + 6.01 + 6.25 + 5.85 + 5.91 + 6.05 + 5.88 + 5.91}{10} = \frac{58.89}{10} = 5.889
\]
where:
- \(n = 10\) (number of bottles),
- \(x_i\) are the individual measurements.
The population standard deviation (\(\sigma\)) is known to be 0.3 ounces, which allows the use of a Z-test for the hypothesis testing.
Calculating the Test Statistic
The test statistic (Z) is calculated using the formula:
\[
Z = \frac{\bar{x} - \mu_{0}}{\sigma / \sqrt{n}}
\]
where:
- \(\bar{x} = 5.889\),
- \(\mu_0 = 6\) (the hypothesized population mean),
- \(\sigma = 0.3\),
- \(n = 10\).
Substituting:
\[
Z = \frac{5.889 - 6}{0.3 / \sqrt{10}} = \frac{-0.111}{0.3 / 3.162} = \frac{-0.111}{0.0949} \approx -1.17
\]
The calculated Z-value is approximately -1.17.
Finding the P-value and Explanation
The P-value for a Z-test in a one-tailed test test at the lower tail (less than 6 ounces) corresponds to the probability of observing a Z-value less than -1.17.
Using a standard normal distribution table or calculator, the P-value is:
\[
P(Z
\]
Thus, the probability of obtaining a sample mean as low as 5.889 or lower, assuming the null hypothesis is true, is about 12.10%.
How the P-value was obtained:
- Calculated by entering Z = -1.17 into a statistical software or calculator capable of computing normal probabilities.
- The result indicates the tail probability to the left of Z = -1.17 in the standard normal distribution.
Conclusion of the Hypothesis Test
Since the P-value (0.1210) exceeds the significance level \(\alpha = 0.05\), we fail to reject the null hypothesis. There is insufficient evidence at the 5% level to conclude that the mean fill of the bottles is less than the labeled 6 ounces. Therefore, based on this sample, the filling process appears to be adequate and consistent with the stated amount.
Confidence Interval for the Mean
Now, moving to the second part concerning the bread company's rye flour usage:
The sample size is \(n=50\), mean \(\bar{x} = 180\) pounds, standard deviation \(s=38.5\).
For a 90% confidence interval, the T-distribution is used because the population standard deviation is unknown and the sample size is concerned with the sample standard deviation.
The formula for the confidence interval:
\[
\bar{x} \pm t_{(1-\alpha/2,\,df)} \times \frac{s}{\sqrt{n}}
\]
where:
- \(df = n - 1 = 49\),
- \(\alpha = 0.10\), so \((1 - \alpha/2) = 0.95\).
Using T-Distribution with calculator function T1-83/84:
- Entered as: `tcdf(49,1E99,tinv(0.95,49))` for finding \(t_{0.975,49}\).
- The critical t-value from the calculator is approximately 1.68.
Calculating the margin of error (ME):
\[
ME = t_{(0.95,49)} \times \frac{s}{\sqrt{n}} = 1.68 \times \frac{38.5}{\sqrt{50}} \approx 1.68 \times 5.45 \approx 9.15
\]
The 90% confidence interval:
\[
180 \pm 9.15
\]
which is:
\[
(170.85, 189.15)
\]
Summary:
- The 90% confidence interval for the mean daily rye flour usage is approximately (170.85, 189.15) pounds.
- The margin of error, rounded to the nearest tenth, is 9.2 pounds.
Conclusion
The interval suggests that the company's average daily rye flour usage is between approximately 171 and 189 pounds with 90% confidence, providing valuable information for inventory planning and resource management.
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