Question

# Please explain how to Solve. Thank you 16. The following are interest rates (annual percentage rates)...

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)) < (Xμ) / (σ/sqrt(n)) < (30−25) /(15/sqrt(36))

= P ( (26 − 25) / (15/6) < (Xμ) / (σ/sqrt(n)) < (30−25) /(6)

Since Z = (xμ) / (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

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