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QUESTION 2

(a) In an experiment of breeding mice, a geneticist has obtained the following results:

- 120 brown mice with pink eyes
- 48 brown mice with brown eyes
- 36 white mice with pink eyes
- 13 white mice with brown eyes

The theory predicts that these types of mice should be obtained with the genetic percentages of 56%, 19%, 19%, and 6%, respectively. Test the compatibility of the data with the theory using a 0.05 level of significance.

(b) Three different shops are used to repair electric motors. One hundred motors are sent to each shop. When a motor is returned, it is classified as either complete, requiring an adjustment, or incomplete repair. Based on the data in Table 4, use a 0.05 level of significance to test whether there is homogeneity among the shops' repair distribution.

Table 4:

| Repair Type | Shop 1 | Shop 2 | Shop 3 |
|---------------|--------|--------|--------|
| Complete | 78 | 56 | 54 |
| Adjustment | 15 | 30 | 31 |
| Incomplete | 7 | 14 | 15 |
| Total | 100 | 100 | 100 |

Answer :

(a) To test the compatibility of data with theory in the breeding mice experiment, we can use the chi-square goodness-of-fit test.

The null hypothesis (H0) is that the observed frequencies are consistent with the expected frequencies based on the theory. The alternative hypothesis (Ha) is that there is a significant difference between the observed and expected frequencies.

The expected frequencies can be calculated by multiplying the total number of mice by the respective genetic percentages. In this case, the expected frequencies are:

Expected frequencies for brown mice with pink eyes: (120+48+36+13) * 0.56 = 150

Expected frequencies for brown mice with brown eyes: (120+48+36+13) * 0.19 = 50

Expected frequencies for white mice with pink eyes: (120+48+36+13) * 0.19 = 50

Expected frequencies for white mice with brown eyes: (120+48+36+13) * 0.06 = 16

Now we can calculate the chi-square test statistic:

χ^2 = Σ((Observed frequency - Expected frequency)^2 / Expected frequency)

Using the given observed frequencies and the calculated expected frequencies, we can calculate the chi-square test statistic. If the test statistic is greater than the critical value from the chi-square distribution table at the chosen level of significance (0.05), we reject the null hypothesis.

(b) To test the homogeneity of repair distribution among the three shops, we can use the chi-square test of independence.

The null hypothesis (H0) is that there is no association between the shop and the type of repair. The alternative hypothesis (Ha) is that there is an association between the shop and the type of repair.

We can construct an observed frequency table based on the given data:

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| Shop 1 | Shop 2 | Shop 3 | Total

Complete | - | 78 | 56 | 134

Adjustment | - | 15 | 30 | 45

Incomplete | - | 7 | 14 | 21

Total | 100 | 100 | 100 | 200

To perform the chi-square test of independence, we calculate the expected frequencies under the assumption of independence. We can calculate the expected frequencies by multiplying the row total and column total for each cell and dividing by the overall total.

Once we have the observed and expected frequencies, we can calculate the chi-square test statistic:

χ^2 = Σ((Observed frequency - Expected frequency)^2 / Expected frequency)

If the test statistic is greater than the critical value from the chi-square distribution table at the chosen level of significance (0.05), we reject the null hypothesis.

Learn more about frequencies here -: brainly.com/question/254161

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