To perform a hypothesis test to determine if the average sodium content for single servings of Special K is less than 220 milligrams at a 0.01 level of significance, we can set up the following hypotheses:
- Null hypothesis (\(H_0\)): \(\mu = 220\) mg (the average sodium content is equal to 220 mg)
- Alternative hypothesis (\(H_a\)): \(\mu < 220\) mg (the average sodium content is less than 220 mg)
Step 1: Calculate the test statistic
We will use a one-sample t-test since the sample size is small (n < 30). The formula for the t-test statistic is:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Where:
- \(\bar{x} = 200\) mg (the sample mean)
- \(\mu_0 = 220\) mg (the population mean under the null hypothesis)
- \(s = 25\) mg (the sample standard deviation)
- \(n = 20\) (the sample size)
Plugging in the values:
\[ t = \frac{200 - 220}{25 / \sqrt{20}} \]
Calculating the denominator:
\[ 25 / \sqrt{20} \approx 25 / 4.472 \approx 5.585 \]
Now substitute back into the equation for \(t\):
\[ t = \frac{-20}{5.585} \approx -3.579 \]
Step 2: Determine the critical value
Next, we need to find the critical t-value for a one-tailed test at the 0.01 significance level with \(n - 1 = 20 - 1 = 19\) degrees of freedom.
Looking at the t-distribution table, the critical value \(t_{0.01, 19}\) is approximately -2.861.
Step 3: Compare the test statistic to the critical value
Now that we have our test statistic of approximately -3.579 and our critical value of -2.861, we can compare them.
- If \(t < -2.861\), we reject the null hypothesis.
- If \(t \geq -2.861\), we do not reject the null hypothesis.
Since -3.579 < -2.861, we reject the null hypothesis.
Conclusion
At the 0.01 level of significance, there is sufficient evidence to suggest that the average sodium content for single servings of Special K is less than 220 mg.
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