In a recent poll of 210 south Jersey households, it was found that 152 households had at least one computer. Construct a 95% confidence interval to estimate the population proportion of south Jersey. A separate survey in Philadelphia found the sample proportion of households with more than one computer to be 0.66, what kind conclusion you can made in terms of the difference in proportion between South Jersey and Philadelphia? (Your conclusion must be based on the confidence interval you obtained.)
confidence interval for population proportion in Jersey
estimated proportion is= 152/210= 0.72 381
zα/2 = 1.9599639861
Lower Bound = p̂ - zα/2•√p̂(1 - p̂)/n = 0.72381 -
(1.9599639861)(0.030853654865695065)
= 0.6633
Upper Bound = p̂ + zα/2•√p̂(1 - p̂)/n
= 0.72381 + (1.9599639861)(0.030853654865695065)
= 0.7843
Confidence Interval = (0.6633, 0.7843)
it means that true population proportion in Jersey is 66.3 percent to 78.43 percent
now in Philadelphia population proportion is 66%. and above confidence interval does not contain 66%. confidence interval is always greater than 66%.
it means that New Jersey has more proportion of people having at least one computer than Philadelphia.
please like ??
Get Answers For Free
Most questions answered within 1 hours.