Write 761/10,000 as a decimal number
1 answer:
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Using the z-distribution, it is found that the 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which z is the z-score that has a p-value of .
195 out of 500 people said they would take more vacations this year than last year, hence:
95% confidence level, hence, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).
A similar problem is given at brainly.com/question/15850972