a) IHI insurance is interested in determining whether younger drivers (below the age of 25) are more prone to motor accidents than older drivers (25 or older). A random sample of 70 IHI policyholders below the age of 25 were selected and whether they had been in less than 2 (or 2 or more) motor accidents in the past year was recorded. This process was repeated for a random sample of 56 IHI policyholders aged 25 or older. The data from these samples is displayed in the table.

a)The IHI statisticians would like to test whether the proportion of policyholders who have had at least 2 accidents in the past year is the same for those who are aged below 25 (<25) and those who are aged 25 or older (¡Ý25). They have constructed a hypothesis test with H0: <25 = ¡Ý25 and H1: <25 ¡Ù ¡Ý25. Calculate the test statistic (¦Ö2) that corresponds to this hypothesis test. Give your answer to 3 decimal places.

Accidents Age Total
< 25 ¡Ý 25
< 2 46 37 83
¡Ý 2 24 19 43
Total 70 56 126

¦Ö2 =

b) Using the test statistic for IHI's hypothesis test and a 95% confidence level, IHI should (accept, reject or retain) the null hypothesis.

We do not have the table, and I am not familiar with some of the symbols you are using.

a) To calculate the test statistic (Χ^2) for this hypothesis test, you need to use the formula:

Χ^2 = Σ [(O - E)^2 / E],

where Σ refers to the sum of all cells, O represents the observed frequencies, and E represents the expected frequencies.

First, let's calculate the expected frequencies. To do this, we'll assume that the null hypothesis is true, which means that the proportion of policyholders with at least 2 accidents is the same for both age groups.

Total number of policyholders below 25: 70
Total number of policyholders 25 or older: 56

Proportion of policyholders below 25 with <2 accidents: 46/70
Proportion of policyholders 25 or older with <2 accidents: 37/56

Expected frequency for <2 accidents below 25: (46/70) * 83
Expected frequency for <2 accidents 25 or older: (37/56) * 83

Proportion of policyholders below 25 with ≥2 accidents: 24/70
Proportion of policyholders 25 or older with ≥2 accidents: 19/56

Expected frequency for ≥2 accidents below 25: (24/70) * 83
Expected frequency for ≥2 accidents 25 or older: (19/56) * 83

Now, we can calculate the test statistic:

Χ^2 = [(46 - E1)^2 / E1] + [(37 - E2)^2 / E2] + [(24 - E3)^2 / E3] + [(19 - E4)^2 / E4],

where E1, E2, E3, and E4 are the expected frequencies for each cell.

Substituting the values:

Χ^2 = [(46 - E1)^2 / E1] + [(37 - E2)^2 / E2] + [(24 - E3)^2 / E3] + [(19 - E4)^2 / E4],

Χ^2 = [(46 - (46/70 * 83))^2 / (46/70 * 83)] + [(37 - (37/56 * 83))^2 / (37/56 * 83)] + [(24 - (24/70 * 83))^2 / (24/70 * 83)] + [(19 - (19/56 * 83))^2 / (19/56 * 83)].

Now, calculate this expression to get the test statistic Χ^2.

b) To determine whether to accept or reject the null hypothesis at a 95% confidence level, you need to compare the calculated test statistic (Χ^2) with the critical value from the chi-square distribution.

The degrees of freedom for this test are (number of rows - 1) * (number of columns - 1). In this case, the degrees of freedom are (2 - 1) * (2 - 1) = 1.

Using a chi-square distribution table or calculator, find the critical value for a 95% confidence level and 1 degree of freedom.

If the calculated test statistic (Χ^2) is greater than the critical value, you would reject the null hypothesis. Otherwise, if it is less than or equal to the critical value, you would retain the null hypothesis.

So, compare the calculated test statistic (Χ^2) from part a) with the critical value at a 95% confidence level (found from the chi-square distribution table), and make your decision based on the comparison.