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

To find the 95% confidence interval for the percentage of viewers who liked the show, we can use the formula:

$CI = \left[ \hat{p} - MOE, \hat{p} + MOE \right]$

where:

• $\hat{p} = \frac{546}{1560} = 0.35$ (proportion of viewers who liked the show)
• $MOE = 0.025318$ (from part b(i)).

Thus, the confidence interval becomes: $CI = \left[ 0.35 - 0.025318, 0.35 + 0.025318 \right] \approx [ 0.3247, 0.3753 ]$

Converting these proportions to percentages: $CI = [32.5\%, 37.5\%]$

The 95% confidence interval for the percentage of viewers who liked the show is approximately [32.5%, 37.5%].

To conduct a hypothesis test to evaluate the executive's claim, we state:

• Null Hypothesis (H₀): $p = 0.40$ (40% of viewers liked the show)
• Alternative Hypothesis (H₁): $p \neq 0.40$ (the proportion of viewers who liked the show is not 40%)

Using our confidence interval from part b(ii), which is approximately [32.5%, 37.5%], we can conclude:

Since the claimed proportion of 40% is not included within the confidence interval, we reject the null hypothesis (H₀). There is insufficient evidence to support the executive's claim that 40% of viewers liked the show.