To determine if the average sodium content for single servings of Special K is greater than 220 milligrams, we can perform a hypothesis test using the following steps:
Step 1: Set the hypotheses
- Null hypothesis (\(H_0\)): The average sodium content is equal to 220 mg, \( \mu = 220 \) mg.
- Alternative hypothesis (\(H_1\)): The average sodium content is greater than 220 mg, \( \mu > 220 \) mg.
Step 2: Determine the significance level
The significance level (\(\alpha\)) is given as 0.1.
Step 3: Calculate the test statistic
We will use a one-sample \( t \)-test since the population standard deviation is unknown and the sample size is small (n < 30). The formula for the t-test statistic is:
\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]
Where:
- \(\bar{x} = 244\) mg (sample mean)
- \(\mu = 220\) mg (population mean under \(H_0\))
- \(s = 25\) mg (sample standard deviation)
- \(n = 20\) (sample size)
Now plug in the values:
\[ t = \frac{244 - 220}{25 / \sqrt{20}} = \frac{24}{25 / \sqrt{20}} = \frac{24 \sqrt{20}}{25} \]
Calculating \( \sqrt{20} \): \[ \sqrt{20} \approx 4.472 \]
Now calculate the test statistic:
\[ t \approx \frac{24 \times 4.472}{25} \approx \frac{107.328}{25} \approx 4.293 \]
Step 4: Determine the critical value
To find the critical \( t \)-value for one-tailed test at \( \alpha = 0.1 \) with \( n - 1 = 20 - 1 = 19 \) degrees of freedom, we can refer to the t-distribution table.
For \( df = 19 \) and \( \alpha = 0.1 \) (one-tailed), the critical value (\( t_{critical} \)) is approximately:
\[ t_{critical} \approx 1.328 \]
Step 5: Make a decision
Now we compare the calculated \( t \)-statistic to the critical \( t \)-value:
- Calculated \( t \) statistic: \( t \approx 4.293 \)
- Critical \( t \): \( t_{critical} \approx 1.328 \)
Since \( 4.293 > 1.328 \), we reject the null hypothesis.
Conclusion
At the 0.1 level of significance, we have sufficient evidence to suggest that the average sodium content for single servings of Special K is greater than 220 milligrams.
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