4. A survey of 320 households with 5 pets each revealed the distribution shown in the table.
Is the result consistent with the hypothesis that dogs(male) and cats(female) are equally probable as pets?

Male and female pets	5 male pets	4 male pets	3 male pets	2 male pets	1 male pets	0 male pets	Total
0 female pets	1 female pets	2 female pets	3 female pets	4 female pets	5 female pets
Number of Families 	18	56	110	88	40	8	320

To test whether the result is consistent with the hypothesis that male pets (dogs) and female pets (cats) are equally probable, we can use a **chi-square goodness-of-fit test**.

### Hypothesis:

- Null Hypothesis \( H_0 \): The probability of having male pets (dogs) and female pets (cats) is equal, i.e., \( p_{\text{male}} = p_{\text{female}} = 0.5 \).
- Alternative Hypothesis \( H_1 \): The probability of having male pets (dogs) and female pets (cats) is not equal.

### Given Data:

| Number of male pets | Number of families (observed) |
|---------------------|-------------------------------|
| 5 male pets         | 18                            |
| 4 male pets         | 56                            |
| 3 male pets         | 110                           |
| 2 male pets         | 88                            |
| 1 male pet          | 40                            |
| 0 male pets         | 8                             |
| **Total**           | 320                           |

### Steps to Solve:

1. **Expected Probabilities**:
   Since we assume that the probability of getting a male or female pet is equal (i.e., \( p = 0.5 \)), the number of male pets in each family can be modeled by a **binomial distribution** with parameters \( n = 5 \) (since each family has 5 pets) and \( p = 0.5 \) (since the probability of having a male pet is 50%).

   The probability of getting \( k \) male pets is given by:
   \[
   P(\text{k male pets}) = \binom{5}{k} \cdot (0.5)^k \cdot (0.5)^{5-k} = \binom{5}{k} \cdot (0.5)^5
   \]
   This formula gives the probability of each possible outcome (from 0 to 5 male pets).

2. **Expected Frequencies**:
   Multiply the probabilities by the total number of families (320) to get the expected number of families for each category.

3. **Chi-Square Calculation**:
   The chi-square statistic is calculated using the formula:
   \[
   \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
   \]
   where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency for each category.

Let me compute the expected frequencies and the chi-square statistic.### Results:

1. **Expected Frequencies** (based on the binomial distribution with \( p = 0.5 \)):
   - 5 male pets: 10 families
   - 4 male pets: 50 families
   - 3 male pets: 100 families
   - 2 male pets: 100 families
   - 1 male pet: 50 families
   - 0 male pets: 10 families

2. **Chi-square statistic**: 11.96

3. **p-value**: 0.0353

### Conclusion:
Since the p-value \( (0.0353) \) is less than the typical significance level \( \alpha = 0.05 \), we **reject the null hypothesis**. This indicates that the observed distribution of male and female pets is not consistent with the hypothesis that male and female pets are equally probable (i.e., \( p = 0.5 \)).

Would you like more details on any part of the process?

### Related Questions:
1. What are the assumptions behind a chi-square test?
2. How does a binomial distribution differ from other probability distributions?
3. What happens if the p-value is very close to 0.05?
4. How can you interpret a higher chi-square statistic?
5. Can a hypothesis be rejected at any significance level?

**Tip**: Always make sure the expected frequency in each category is large enough (at least 5) for the chi-square approximation to be valid.

A survey of 320 households with 5 pets each revealed the distribution shown in the table. Is the result consistent with the hypothesis that dogs (male) and cats (female) are equally probable as pets?

Analyze the pet distribution in 320 households using the Chi-Square Goodness-of-Fit Test. This problem explores the hypothesis that male pets (dogs) and female pets (cats) are equally probable, using binomial distribution and chi-square statistics. Suitable for students in Grades 11-12 and college-level statistics courses.

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