statistics
posted by Kee .
In Los Angeles there are three network television stations, each with it own evening news program from 6:00 to 6:30 PM. According to a report in this morning’s local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on KNBC (channel 2), 64 watched KABC (channel 7), and 33 viewed KCBS (channel 2). At the 0.05 significance level, is there a difference in the proportion of viewers watching the three channels?
1. State the null and alternative hypothesis.
2. Identify the level of significance: Given at 0.05

You can probably use a chisquare for this kind of problem.
Hypotheses:
Ho: There was equal distribution of viewers across the channels.
Ha: There was unequal distribution of viewers across the channels.
If there was equal distribution across the channels, you would expect 50 viewers watching each channel (150/3 = 50).
You can use the chisquare formula to calculate the test statistic if you need to do so.
Once you have the chisquare value calculated, compare this test statistic with a value from the chisquare table using df = 2 (df = number of groups minus 1) and .05 level of significance.
If the test statistic calculated exceeds the value you find from the table, the null will be rejected (there is a difference). If the test statistic does not exceed the critical value from the table, the null will not be rejected (there is no difference).
I hope this will help.
Respond to this Question
Similar Questions

please read my introduction for media bias.
This is my introduction to my essay for media bias in television news. The primary source of political information in America is the media. [Media bias is a way to represent different people in a certain way based on their own views.] … 
Statistics
In Los Angeles there are three network television stations, each with it own evening news program from 6:00 to 6:30 PM. According to a report in this morning’s local newspaper, a random sample of 150 viewers last night revealed 53 … 
statistics
In a survey of 5100 TV viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 95% confidence in our estimate of the percent of TV viewers who watch network news programs. Hint, find … 
Math Data management
A survey of 1000 television viewers conducted by a local television station produced the following data: •40% watch the news at 12:00 •60% watch the news at 18:00 •50% watch the news at 23:00 •25% watch the news at 12:00 and … 
statistic homework
If 55% of a television viewing population watched a program called ¡§Name that Poem¡¨ one evening, what is the probability that, in a random sample of 100 viewers, less than 50% of the sample watched the program? 
statistic
If 55% of a television viewing population watched a program called ¡§Name that Poem¡¨ one evening, what is the probability that, in a random sample of 100 viewers, less than 50% of the sample watched the program? 
Statistics
please help having difficulty with this problem. It was noted that 40% of the people in Dallas read the morning newspaper. A random sample of 18 people is selected from the city. On average, how many people would you expect to read … 
Data Management Grade 12
A survey of 1000 television viewers conducted by a local television station produced the following data: •40% watch the news at 12:00 •60% watch the news at 18:00 •50% watch the news at 23:00 •25% watch the news at 12:00 and … 
econ 112
A random sample of 400 morning shoppers showed that 130 were men. A random sample of 480 evening shoppers showed 187 to be men. Use a 5% level of significance to test for a significant difference in the proportion of morning and evening … 
prelag8
The AllInOne cable TV network takes a random sample of viewers that watch the network to get their opinion on a new TV show. They then advertise the following: “Ninety percent of all viewers rate our new show as the best they have …