Please help it has to be in fraction form rounded all the way
1 answer:
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Answer:
The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).
Step-by-step explanation:
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 zscore that has a pvalue of .
For this problem, we have that:
92% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).