fifty two of a test panel of 175 consumers say that they would buy cereal A if it is presented on the market and 35 of another test panel of 150 consumers say that they would buy cereal B. Test the null hypothesis that the cereals on the market would be equally popular. Use a 5% significance level. Give the null and alternative
The null hypothesis (H₀): The cereals on the market are equally popular.
The alternative hypothesis (H₁): The cereals on the market are not equally popular.
To solve this problem, we will perform a chi-square test of independence. We will create a contingency table with the observed frequencies of consumers who would buy cereal A and cereal B.
| Cereal A | Cereal B | Total
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Buy | 52 | 35 | 87
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Not Buy | 123 | 115 | 238
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Total | 175 | 150 | 325
Now, let's calculate the expected frequencies in each cell of the contingency table under the assumption that the cereals are equally popular.
Expected frequency for Cereal A and Buy = (number of Cereal A buyers * total number of buyers) / total number of consumers
Expected frequency for Cereal B and Buy = (number of Cereal B buyers * total number of buyers) / total number of consumers
Expected frequency for Cereal A and Not Buy = (number of Cereal A non-buyers * total number of non-buyers) / total number of consumers
Expected frequency for Cereal B and Not Buy = (number of Cereal B non-buyers * total number of non-buyers) / total number of consumers
Expected frequency for Cereal A and Buy = (175 * 87) / 325 ≈ 46.215
Expected frequency for Cereal B and Buy = (150 * 87) / 325 ≈ 40.200
Expected frequency for Cereal A and Not Buy = (175 * 238) / 325 ≈ 128.538
Expected frequency for Cereal B and Not Buy = (150 * 238) / 325 ≈ 110.462
Now, we can calculate the chi-square test statistic:
χ² = Σ[(Observed frequency - Expected frequency)² / Expected frequency]
χ² = [(52 - 46.215)² / 46.215] + [(35 - 40.200)² / 40.200] + [(123 - 128.538)² / 128.538] + [(115 - 110.462)² / 110.462]
χ² ≈ 0.514 + 0.643 + 0.193 + 0.201 ≈ 1.551
To determine whether the chi-square statistic is statistically significant, we need to compare it to the critical value from the chi-square distribution with (number of rows - 1) * (number of columns - 1) degrees of freedom at the given significance level (α = 0.05).
In this case, the contingency table has (2 - 1) * (2 - 1) = 1 degree of freedom.
The critical value for α = 0.05 and 1 degree of freedom is approximately 3.841.
Since the calculated chi-square statistic (1.551) is less than the critical value (3.841), we fail to reject the null hypothesis.
Therefore, we do not have enough evidence to conclude that the cereals on the market are not equally popular at a 5% significance level.
The null hypothesis (H0): The cereals on the market are equally popular.
The alternative hypothesis (H1): The cereals on the market are not equally popular.
The null hypothesis, denoted as H0, is the statement we want to test and assumes there is no significant difference between the popularity of cereal A and cereal B on the market. It suggests that both cereals are equally popular.
The alternative hypothesis, denoted as Ha or H1, is the statement that contradicts the null hypothesis. In this case, the alternative hypothesis is that there is a significant difference in the popularity of cereal A and cereal B on the market.
To test the null hypothesis, we need to conduct a hypothesis test using statistical methods. In this case, we will perform a hypothesis test for proportions.
Let's denote the proportion of consumers who would buy cereal A as p1 and the proportion of consumers who would buy cereal B as p2. We will compare these proportions to see if they are significantly different.
The test statistic we will use is called the z-score. The formula for calculating the z-score is:
z = (p1 - p2) / sqrt[(p * (1-p)/n1) + (p * (1-p)/n2)]
Where:
- p is the pooled proportion, calculated as (p1 * n1 + p2 * n2) / (n1 + n2)
- n1 and n2 are the sample sizes (175 and 150 in this case)
- p1 and p2 are the sample proportions (52/175 and 35/150)
The next step is to determine the critical value to compare the test statistic. We will choose a 5% significance level, which corresponds to a 95% confidence level. The critical value for a two-tailed test at this significance level is approximately ±1.96.
If the calculated z-score falls outside the range of -1.96 to +1.96, we reject the null hypothesis in favor of the alternative hypothesis. Otherwise, if the calculated z-score falls within this range, we fail to reject the null hypothesis.
To summarize:
- Null hypothesis (H0): The popularity of cereal A and cereal B on the market is equal.
- Alternative hypothesis (Ha): The popularity of cereal A and cereal B on the market is significantly different.
Now we can proceed to calculate the test statistic (z-score) using the provided sample data.