Answer:
0.4524, 0.5076), Not necessary, not supported
Explanation:
a)Since 48% is the sample proportion for a sample of 1259 US residents, it is a characteristic of a sample hence it is a statistic
b) For 95% confidence interval first we calculate std error
STd error =
![√(pq/n) \\\\=0.0141](https://img.qammunity.org/2020/formulas/mathematics/college/w772jizrtl1d5nl9oemqpe8656c4ospbb7.png)
Margin of error for 95% = ±1.96*0.0141=0.0276
Confidence interval 95% = (0.48±0.0276)
=(0.4524, 0.5076)
We are 95% confidence that for samples of large size randomly drawn, the sample proportion falls within this range inside the interval
c) No, it need not be normal. The conditions are only
samples should be randomly drawn, and sample size should be sufficiently large to represent the population
d) We find that the confidence interval contains 0.50 and also mid value below 0.50
Hence the confidence interval does not support the news piece statement because it is evident only 50% support and not more than 50%