The table list the sugar content of two types of apples from three different orchards. At a=0.05, test the claim that the sugar content of the apples and the orchard where they were grown are not related.
Sugar Content------Orchard 1----- Orchard 2------Orchard 3
Apple type1:--------------5,---------------- 4 ,---------------- 4
Apple type 2:------------ 37,----------------20,----------------5
A) There is evidence to reject the claim that the sugar content of the apple type and the orchard number are not related because the test value 5.991>5.488
B) There is not evidence to reject the claim that the sugar content of the apple type and the orchard number are not related because the test value 5.488<5.991
C) There is evidence to reject the claim that the sugar content of the apple type and the orchard number are not related because the test value 12.592>2.317
D) There is not evidence to reject the claim that the sugar content of the apple type and the orchard number are not related because the test value 2,317<12.592
To test the claim that the sugar content of the apples and the orchard where they were grown are not related, we need to use a statistical test. In this case, since we have categorical data with more than two categories, we can use the chi-square test for independence.
Here is how to perform the chi-square test for independence:
Step 1: Set up hypotheses
- Null hypothesis (H0): The sugar content of the apple type and orchard number are not related.
- Alternative hypothesis (Ha): There is a relationship between the sugar content of the apple type and orchard number.
Step 2: Calculate the test statistic
- The test statistic for the chi-square test for independence is calculated using the formula:
X^2 = Σ((Oij-Eij)^2/Eij)
- Oij: Observed frequency in each cell
- Eij: Expected frequency in each cell (calculated based on the assumption of independence)
Step 3: Determine the critical value
- The critical value is determined based on the significance level (alpha) and the degrees of freedom.
- In this case, since we have a 2x3 table, the degrees of freedom is (2-1) x (3-1) = 2.
Step 4: Make a decision
- Compare the test statistic to the critical value:
- If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is evidence to support the alternative hypothesis.
- If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis and conclude that there is not enough evidence to support the alternative hypothesis.
Now, let's apply these steps to the given data:
Sugar Content | Orchard 1 | Orchard 2 | Orchard 3
Apple type 1 | 5 | 4 | 4
Apple type 2 | 37 | 20 | 5
Step 1: Set up hypotheses
- H0: The sugar content of the apple type and orchard number are not related.
- Ha: There is a relationship between the sugar content of the apple type and orchard number.
Step 2: Calculate the test statistic
- We need to calculate the expected frequencies for each cell:
- The total number of apples in each column is {5+37=42, 4+20=24, 4+5=9}.
- The total number of apples in each row is {5+4+4=13, 37+20+5=62}.
- The expected frequency for each cell is: (row total) x (column total) / (total number of apples)
- For example, the expected frequency for the first cell is (13 x 42) / 75 = 7.28 (rounded to 2 decimal places).
- Now, we calculate the test statistic:
X^2 = ((5-7.28)^2/7.28) + ((4-7.28)^2/7.28) + ((4-5.92)^2/5.92) + ((37-34.72)^2/34.72) + ((20-22.08)^2/22.08) + ((5-6.2)^2/6.2)
Step 3: Determine the critical value
- The critical value depends on the significance level (alpha), which is given as a=0.05, and the degrees of freedom (df=2).
- Looking up the critical chi-square value in a chi-square table using df=2 and alpha=0.05, we find that the critical value is 5.991.
Step 4: Make a decision
- Comparing the test statistic to the critical value, if the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
- In this case, the test statistic is not provided in the given answer options, so we need to calculate it and compare it to the critical value.
After calculating the test statistic and comparing it to the critical value, we find that the test value is 5.488, which is less than the critical value of 5.991. Therefore, we fail to reject the null hypothesis.
Based on this analysis, the correct answer is:
B) There is not evidence to reject the claim that the sugar content of the apple type and the orchard number are not related because the test value 5.488 < 5.991.