During the first 13 weeks of the television season, the Saturday evening 8:00 P. M. to 9:00 P. M. audience proportions were recorded as ABC 29%, CBS 28%, NBC 25%, and independents 18%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following viewing audience data: ABC 95 homes, CBS 65 homes, NBC 87 homes, and independents 53 homes. Test with = .05 to determine whether the viewing audience proportions changed. Required:
Compute the value of the 2 test statistic (to 2 decimals). The p value is Has a significant change occurred in viewing audience proportions?

- The value of the 2 test statistic is 6.98

- There is no sufficient evidence to reject the claim of the specific distribution

Step-by-step explanation:

Given

Proportion (p)

Samples

First, we formulate a table for chi square totals:

Where

O = Observed Frequency and E = Expected Frequency

So, we have:

For ABC

For CBS

For NBC

For others:

--- approximated

The value of the 2 test statistic is 6.98

Next, is to determine the p value:

Calculate the degree of freedom

Using the table in the 3rd row,

corresponds to:

The value of p is:

This implies that, we fail to reject H o

x^5/3y^1/3 the correct answer

hope it

answer: procent 54 wynosi 33,33%