IWHDHDNDJBFJDNX HELP!!!!!
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Answer:
The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).
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:
90% 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 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).
Answer:
0.998 is the probability that the average money spent by a sample of 40 shoppers is within $10 of the actual population mean.
Step-by-step explanation:
We are given the following information in the question:
Standard Deviation, σ = $21.51
We are given that the distribution of average money spend is a bell shaped distribution that is a normal distribution.
Formula:
We have to find:
P( average money spent is within $10 of the actual population mean.)
Calculation the value from standard normal z table, we have,