Charles wants to find out if the students in foreign language classes spend more time in class speaking in English or in the for
1 answer:
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<h2>Answer with explanation:</h2>
The confidence interval for population mean is given by :-
, where is the sample proportion, n is the sample size ,
is the critical z-value.
The values needed to calculate a confidence interval at the 99% confidence level are :
Given : Significance level :
Sample size : n=450
Critical value :
Sample proportion:
Now, the 99% confidence level will be :
Hence, the 99% confidence interval is
Answer:
- The lowest value of the confidence interval is 0.5262 or 52.62%
- The highest value of the confidence interval is 0.5538 or 55.38%
Step-by-step explanation:
Here you estimate the proportion of people in the population that said did not have children under 18 living at home.It can also be given as a percentage.
The general expression to apply here is;
where ;
p=sample proportion
n=sample size
z*=value of z* from the standard normal distribution for 95% confidence level
Given;
n=5000
<u>Find p</u>
From the question 54% of people chosen said they did not have children under 18 living at home
<u>To calculate the 95% confidence interval, follow the steps below;</u>
- Find the value of z* from the z*-value table
The value of z* from the table is 1.96
- calculate the sample proportion p
The value of p=0.54 as calculated above
- Find p(1-p)
- Find p(1-p)/n
Divide the value of p(1-p) with the sample size, n
- Find the square-root of p(1-p)/n
- Find the margin of error
Here multiply the square-root of p(1-p)/n by the z*
The 95% confidence interval for the lower end value is p-margin of error
The 95% confidence interval for the upper end value is p+margin of error