What is the solution of this equqtion? 1w= -17
2 answers:
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Answer:
A. Normal
B. Between 2.078 and 2.722 visits.
C. About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
x% confidence interval:
A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.
Question A:
By the Central Limit Theorem, normal.
Question B:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
That is z with a pvalue of , so Z = 1.645.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 2.4 - 0.322 = 2.078 visits
The upper end of the interval is the sample mean added to M. So it is 2.4 + 0.322 = 2.722 visits.
Between 2.078 and 2.722 visits.
Question c:
90% confidence level, so 90% will contain the true population mean, 100 - 90 = 10% wont.
About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.