According to “Dietary Goals for the United States (1997)”, high sodium intake may be related to ulcers, stomach cancer, and migraine headaches. The human requirement for salt is only 220 milligram per day, which is surpassed in most single servings of ready-to-eat cereals. If a random sample of 20 similar servings of Special K has a mean sodium content of 244 milligrams and a standard deviation 24,5 milligrams does this suggest at 0,01 level of significance that the average sodium content for single servings of Special K is greater than 220 milligrams? Assume the distribution of sodium content to be normal. The value of the test statistic for testing this hypothesis is ____
To test whether the average sodium content for single servings of Special K is greater than 220 milligrams, we can use a one-sample t-test.
The null hypothesis (H0) is that the average sodium content is equal to or less than 220 milligrams.
The alternative hypothesis (Ha) is that the average sodium content is greater than 220 milligrams.
The test statistic can be calculated using the formula:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
Let's calculate the test statistic:
Sample mean = 244 milligrams
Hypothesized mean = 220 milligrams
Standard deviation = 24.5 milligrams
Sample size (n) = 20
t = (244 - 220) / (24.5 / sqrt(20))
t ≈ 5.2
To find the critical value for a 0.01 level of significance, we need to use a t-distribution table with degrees of freedom equal to (n-1) = (20-1) = 19.
Looking up the critical value for a one-tailed test with 0.01 level of significance and 19 degrees of freedom, we find the critical value to be approximately 2.861.
Since the test statistic (t = 5.2) is greater than the critical value (2.861), we reject the null hypothesis. This suggests that at a significance level of 0.01, the average sodium content for single servings of Special K is significantly greater than 220 milligrams.
To determine the value of the test statistic for testing this hypothesis, we need to perform a one-sample t-test. Here are the steps to calculate the test statistic:
Step 1: State the null hypothesis (H0) and the alternative hypothesis (Ha).
In this case, the null hypothesis is that the average sodium content for single servings of Special K is equal to or less than 220 milligrams (μ ≤ 220). The alternative hypothesis is that the average sodium content is greater than 220 milligrams (μ > 220).
Step 2: Set the significance level (α).
The significance level, denoted as α, is given as 0.01 in the question. This represents the probability of making a Type I error (rejecting the null hypothesis when it is true).
Step 3: Compute the test statistic.
The formula for the test statistic in a one-sample t-test is:
t = (x̄ - μ) / (s / √n)
Where:
x̄ is the sample mean (244 milligrams),
μ is the hypothesized population mean (220 milligrams),
s is the sample standard deviation (24.5 milligrams),
n is the sample size (20).
Plugging in the values, we get:
t = (244 - 220) / (24.5 / √20)
t = 24 / (24.5 / 4.47)
t ≈ 8.82
Step 4: Determine the critical value and compare it with the test statistic.
Since we assumed the distribution to be normal and the sample size is relatively small (< 30), we will use the t-distribution. At a significance level of 0.01 and degrees of freedom (df) equal to n - 1 = 20 - 1 = 19, the critical t-value can be found using statistical tables or a t-distribution calculator.
For a one-tailed test (since we're testing if the average sodium content is greater than 220), the critical t-value is approximately 2.861.
Step 5: Make a decision.
If the absolute value of the test statistic is greater than the critical t-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the test statistic (t = 8.82) is greater than the critical t-value (2.861). Therefore, we reject the null hypothesis.
The value of the test statistic for testing this hypothesis is 8.82.
To determine if the average sodium content for single servings of Special K is greater than 220 milligrams, we can perform a one-sample t-test.
Given:
Sample mean (x̄) = 244 milligrams
Standard deviation (σ) = 24.5 milligrams
Sample size (n) = 20
Hypothesized mean (μ) = 220 milligrams
Significance level = 0.01
Step 1: Set up the hypotheses.
The null hypothesis (H0) assumes that the average sodium content is not greater than 220 milligrams.
H0: μ ≤ 220
The alternative hypothesis (Ha) assumes that the average sodium content is greater than 220 milligrams.
Ha: μ > 220
Step 2: Compute the test statistic.
The test statistic for a one-sample t-test is calculated as follows:
t = (x̄ - μ) / (σ / sqrt(n))
Substituting the given values:
t = (244 - 220) / (24.5 / sqrt(20))
Step 3: Determine the critical value.
Since the significance level is 0.01 and the test is one-tailed (right-tailed), we need to find the critical value using a t-table or a t-distribution calculator with the degrees of freedom (df) equal to (n - 1) = (20 - 1) = 19.
The critical value corresponding to a significance level of 0.01 and df = 19 is approximately 2.539.
Step 4: Compare the test statistic with the critical value.
If the calculated test statistic is greater than the critical value, we reject the null hypothesis in favor of the alternative hypothesis. Otherwise, we fail to reject the null hypothesis.
Comparing the calculated test statistic to the critical value:
t = (244 - 220) / (24.5 / sqrt(20))
t ≈ 1.973
Since 1.973 < 2.539, we fail to reject the null hypothesis.
Step 5: State the conclusion.
Based on the given data and the calculated test statistic, at a 0.01 level of significance, there is not enough evidence to suggest that the average sodium content for single servings of Special K is greater than 220 milligrams.