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A survey was carried out on behalf of a television station to investigate the popularity of a certain show - Leaving Cert Mathematics - Question b - 2018
Question b
A survey was carried out on behalf of a television station to investigate the popularity of a certain show. (i) A random sample of 1560 television viewers was surve... show full transcript
Worked Solution & Example Answer:A survey was carried out on behalf of a television station to investigate the popularity of a certain show - Leaving Cert Mathematics - Question b - 2018
Step 1
Find the margin of error of the survey.
Answer
To find the margin of error (ME), we use the formula:
ME = rac{1}{ ext{sqrt{n}}}
where n is the sample size (1560). Thus,
= 0.025318$$ To express the margin of error as a percentage, we multiply by 100: $$ME = 0.025318 \times 100 = 2.5\%$$ Hence, the margin of error is 2.5%.Step 2
Use your answer to part b(i) above to create a 95% confidence interval for the percentage of viewers who liked the show.
Answer
From part (i), we have the margin of error as 2.5%. Now, the proportion of viewers who liked the show is given by:
The 95% confidence interval (CI) is calculated as:
This gives:
Therefore, the 95% confidence interval for the percentage of viewers who liked the show is [32.5%, 37.5%].
Step 3
State your null hypothesis, your alternative hypothesis and give your conclusion in the context of the question.
Answer
Let:
- Null hypothesis (): (40% of viewers liked the show)
- Alternative hypothesis (): (the percentage of viewers who liked the show does not equal 40%)
From part (ii), the confidence interval [32.5%, 37.5%] does not contain 40%. Therefore, we reject the null hypothesis at the 5% level of significance.
In conclusion, there is sufficient evidence to suggest that the percentage of viewers who liked the show does not equal 40%.