Politicians are interested in knowing the opinions of their constituents on important issues. One administrative assistant to a senator claims that more than 63% of adult women favor stricter gun laws. A recent telephone survey of 1026 adult women by IBR Polls found that 65.9% of adult women favored stricter gun laws.
The 90% confidence interval is: ( % , %)
Enter each number as a percent rounded to tenths of a percent.
I got 63.5 to 68.3. Is my answer correct?
n = 1026
za/2 = 1.645
phat = .659
qhat = 1-phat
qhat = 1-.659 = .341
Error = z/2* sqrt( phat *qhat/n)
E = 1.645 * sqrt(.659 * .341/1026))
E = 0.0243
Phat -E <p< phat + E
(.659-.0243, .659+ .0243)
(.6347, .6833) or ( 63.5% , 68.3%)
The 90% confidence interval is : (63.5%, 68.3%)
Is the p-value 0.4995?
To calculate the 90% confidence interval for the proportion of adult women who favor stricter gun laws based on the information given, we can use the formula:
CI = p̂ ± Z * √((p̂(1-p̂))/n)
Where:
- p̂ is the sample proportion (65.9% = 0.659)
- Z is the Z-score for a 90% confidence level (critical value) - which corresponds to a 1.645 Z-score for a one-tailed test
- n is the sample size (1026)
Let's plug in the values:
CI = 0.659 ± 1.645 * √((0.659*(1-0.659))/1026)
CI = 0.659 ± 0.016
Now, to find the actual confidence interval, we need to add and subtract the result above from the sample proportion:
CI = (0.659 - 0.016, 0.659 + 0.016)
CI = (0.643, 0.675)
Now, rounding to the nearest tenth of a percent, we get:
90% confidence interval = (64.3%, 67.5%)
Hence, your answer of 63.5% to 68.3% is incorrect. The correct 90% confidence interval for the proportion of adult women who favor stricter gun laws based on the given data should be (64.3%, 67.5%).