High School

LLD Records is a store that specializes in old-fashioned vinyl records. The store manager collected intent-to-purchase scores for a newly acquired vinyl record from 115 college students across different channels. The scores range from 1 (no intent to purchase) to 100 (full intent to purchase).

The information is as follows:

- **On Campus**:
- Sample Size: 23
- Sample Mean: 59.1
- Sample Variance: 52.8

- **By Email**:
- Sample Size: 23
- Sample Mean: 65.4
- Sample Variance: 92.6

- **By Phone**:
- Sample Size: 23
- Sample Mean: 68.2
- Sample Variance: 117.0

- **On the Website**:
- Sample Size: 23
- Sample Mean: 63.7
- Sample Variance: 59.9

- **In the Store**:
- Sample Size: 23
- Sample Mean: 59.1
- Sample Variance: 78.1

Suppose the five populations of scores from which these samples were drawn are approximately normally distributed and have the same mean and the same variance.

**Task**: Estimate the common population variance by pooling the sample variances given.

Carry your intermediate computations to at least three decimal places and round your final response to at least one decimal place.

Answer :

The estimate of the common population variance is approximately 80.1.

To estimate the common population variance, we can pool the sample variances using the formula for the pooled sample variance:

[tex]\[ s_p^2 = \frac{{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2 + \cdots + (n_k - 1)s_k^2}}{{n_1 + n_2 + \cdots + n_k - k}} \][/tex]

Where:

-[tex]\( s_p^2 \)[/tex] is the pooled sample variance,

[tex]- \( n_i \)[/tex] is the sample size of the ( i )-th group,

-[tex]\( s_i^2 \)[/tex] is the sample variance of the ( i )-th group,

- k is the number of groups.

Substituting the given values:

[tex]\[ s_p^2 = \frac{{(23 - 1)(52.8) + (23 - 1)(92.6) + (23 - 1)(117.0) + (23 - 1)(59.9) + (23 - 1)(78.1)}}{{23 + 23 + 23 + 23 + 23 - 5}} \]\[ s_p^2 = \frac{{22(52.8) + 22(92.6) + 22(117.0) + 22(59.9) + 22(78.1)}}{{115 - 5}} \]\[ s_p^2 = \frac{{1161.6 + 2037.2 + 2574.0 + 1317.8 + 1718.2}}{{110}} \]\[ s_p^2 = \frac{{8808.8}}{{110}} \]\[ s_p^2 = 80.080 \][/tex]

Rounding to one decimal place, the estimate of the common population variance is approximately 80.1.

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