Question

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.

Answer 1

	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