Answer:
Explanation:
(i) The level of significance (alpha) for a 95% confidence interval is 0.05.
(ii) The corresponding Z-value for a 95% confidence interval can be found in the Z-table or using a calculator, and is 1.96.
(iii) The sample proportion of people who preferred seafood is:
p-hat = 160/500 = 0.32
(iv) The formula to calculate the margin of error (ME) for a proportion is:
ME = Z * sqrt((p-hat * (1 - p-hat)) / n)
where Z is the Z-value for the desired level of confidence, p-hat is the sample proportion, and n is the sample size.
(v) Plugging in the values, we get:
ME = 1.96 * sqrt((0.32 * (1 - 0.32)) / 500)
ME ≈ 0.045
So the margin of error is approximately 0.045.
To find the 95% confidence interval, we use the formula:
CI = p-hat ± ME
Plugging in the values, we get:
CI = 0.32 ± 0.045
CI = (0.275, 0.365)
Therefore, we can say with 95% confidence that the actual proportion of people who prefer seafood on Fridays is between 0.275 and 0.365.
(vi) To find the length of the interval, we subtract the lower limit from the upper limit:
Length of interval = 0.365 - 0.275 = 0.09
For a 90% confidence interval, the Z-value is 1.65. Using the same formula and calculations, we get:
ME = 1.65 * sqrt((0.32 * (1 - 0.32)) / 500)
ME ≈ 0.039
CI = 0.32 ± 0.039
CI = (0.281, 0.359)
Length of interval = 0.359 - 0.281 = 0.078
For a 99% confidence interval, the Z-value is 2.58. Using the same formula and calculations, we get:
ME = 2.58 * sqrt((0.32 * (1 - 0.32)) / 500)
ME ≈ 0.057
CI = 0.32 ± 0.057
CI = (0.263, 0.377)
Length of interval = 0.377 - 0.263 = 0.114
For a 99.9% confidence interval, the Z-value is 3.291. Using the same formula and calculations, we get:
ME = 3.291 * sqrt((0.32 * (1 - 0.32)) / 500)
ME ≈ 0.078
CI = 0.32 ± 0.078
CI = (0.242, 0.398)
Length of interval = 0.398 - 0.242 = 0.156
In conclusion, as the confidence level increases, the length of the confidence interval also increases. This makes sense since as we become more certain about our estimate, we need to allow for a wider range of possible values.