Answer:
a) The 95% confidence interval would be given (0.0702;0.1098).
We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)
b) Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
represent the real population proportion of interest
represent the estimated proportion for the sample
n=800 is the sample size required
represent the critical value for the margin of error
Confidence =0.95 or 95%
Part a
The population proportion have the following distribution
The confidence interval would be given by this formula
For the 95% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
The margin of error is given by :
And replacing into the confidence interval formula we got:
And the 95% confidence interval would be given (0.0702;0.1098).
We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)
Part b
Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.
