Please explain how to Solve. Thank you
16. The following are interest rates (annual percentage rates) for a 30-year-fixed-rate mortgage from a sample of lenders in a certain city. It is reasonable to assume that the population is approximately normal. 4.327, 4.461, 4.547, 4.813, 4.365, 4.772, 4.842. Find the upper bound of the 99% confidence interval for the mean rate. Round three decimal places.
4. A survey of high school students revealed that the number of soft drinks consumed per month was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probability that the average number of soft drinks consumed per month for the sampling was between 26.0 and 30 soft drinks (Round 4 decimal places ex. 0.0048)
6. According to one pollster, 59% of children are afraid of the dark. Suppose that a sample size of 30 is drawn. Find the value of p^, the standard deviation of the distribution of sample proportions. Round to two decimal places.
x | (x-xbar)^2 | |||
4.327 | 0.068944 | Mean(x)=xbar=sum(x)/n | 4.589571 | |
4.461 | 0.016531 | standard deviation(s)=sum(x-xbar)^2/n-1 | 0.217807 | |
4.547 | 0.001812 | n | 7 | |
4.813 | 0.04992 | for 99% confidence level with degree of freedom (n-1)=13 | ||
4.365 | 0.050432 | =1-c%=1-0.99=0.01 | 0.01 | |
4.772 | 0.03328 | degrres of freedom | 6 | |
4.842 | 0.06372 | tc=critical value obtain using t-table with corresponding df=n-1 | 3.707428 | |
Sum | 32.127 | 0.28464 | Margin of error =tc*s/sqrt(n) | 0.305208 |
LCL=xbar-ME | 4.284363 | |||
UCL=xbar+ME | 4.894779 |
#upper bound of the 99% confidence interval for the mean rate
Upper bound=xbar+ME=4.895
Ans2:
Since μ = 25 and σ = 15 we have:
P ( 26 < X < 30 ) =
= P ( (26 − 25) / (15/sqrt(36)) < (X−μ) / (σ/sqrt(n)) < (30−25) /(15/sqrt(36))
= P ( (26 − 25) / (15/6) < (X−μ) / (σ/sqrt(n)) < (30−25) /(6)
Since Z = (x−μ) / (σ/sqrt(n))
P ( 28.9 < X < 30 ) = P ( 1.56 < Z < 2 )
Use the standard normal table to conclude that:
P ( 1.56 < Z < 2 ) = P ( Z < 2 ) − P (Z < 0.4 )
= 0.9772 - 0.6554
= 0.3218
# probabilty thataverage number of soft drinks consumed per month for the sampling was between 26.0 and 30 soft drinks is 0.3218
Part3>.
p^=0.59
1-p^=0.41
n=30
standard deviation = [p( 1 - p ) / n] = [(0.59*0.41) /30 ] = 0.0898
#standard deviation =0.09
Please explain how to Solve. Thank you 16. The following are interest rates (annual percentage rates)...
Question 16 (5 points) Following are interest rates (annual percentage rates) for a 30-year-fixed-rate mortgage from a sample of lenders in a certain city. It is reasonable to assume that the population is approximately normal. 4.327, 4.461, 4.547, 4.324, 4.365, 4.326 , 4.842 GP Find the upper bound of the 99% confidence interval for the mean rate.
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