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5_hrs_solutions.docxProblem 15.10:
a) What is the standard deviation of Juan’s mean result:
We know that the standard deviation of mean related to standard deviation of distribution by
σmean = standard deviation/ square root (n) ……..(1)
Given, n=4 and standard deviation = 10,
So we have σmean =10/sqrt(4) = 10/2 = 5
b) Standard deviation of sample mean = 2, find n
From (1) we have 2=10/sqrt(n)
Sqrt(n) = 10/2
= 5
n =25
When there is a single measurement, it may not be close to the population mean. When averaging the
measurements, it gives a result which will be nearer to the population mean.
Problem 15.12:
a) n=50. Since the sample size is greater than 30 we may use central limit theorem.
We are given, mean = 0.5 and standard deviation = 0.7
We have mean =0.5
Also, we have from (1) we have
σmean = standard deviation/ square root (n) = 0.7/sqrt(50)
= 0.9899
b) Prob( average no of moths in 50 traps is greater than 0.6)
Sample mean = 0.6 population mean = 0.5 standard deviation = 0.7
We have Z = (sample mean – population mean)/(σ/sqrt(n)
Z = (0.6 – 0.5)/(0.7/sqrt(50))
=1.01
P(Z>1.01) = 0.1562
Problem 15.28:
We are given that the scores follow normal distribution.
We have mean µ=25 and standard deviation = 6.5
a) Probability (score of a student is between 20 and 30)
We now standardize the scores.
P((20-25/(6.5/sqrt(1))<Z<P((30-25/(6.5/sqrt(1))
=P(- 0.7692 < Z < 0.7692) =
Rounding of 0.7692, we get 0.77
We need P( - 0.77 < Z < 0) to P(0 < Z < 0.77)
From normal tables we have P( - 0.77 < Z < 0) = 0.2794 = P(0 < Z < 0.77)
=0.2794*2 = 0.5588 = 0.56 .
b) n=25….sampling distribution of x bar
sampling distribution of x bar = sigma / sqrt(n) = 6.5/sqrt(25) = 1.3
c) If n= 25, we need to find P((20-25/(6.5/sqrt(25))<Z<P((30-25/(6.5/sqrt(25))
P( - 5 / (6.5/5) < Z < 5 / (6.5/5))
P( -3.84 < Z < 3.84)
= 0.9999
Problem 16.6:
First we find the sample mean of the given measurements.
=(10.08+9.89+10.05+10.16+10.21+10.11)/6 =10.0833
We have for 90% confidence, Z score to be Z = 1.645, σ = 0.1(given) n= 6
The confidence interval for true conductivity is given by
Sample mean ± Standard error*Z
10.0833 ± 1.645*(0.1/sqrt(6) = (10.01586,10.15075) is the confidence interval for true conductivity.
17.6) The average income of American women who work full-time and have only a high school degree is $31,666. You wonder whether the mean income of female graduates from your local high school who work full-time but have only a high school degree is different from the national average. You obtain income information from an SRS of 62 females graduates who work full-time and have only a high school degree and find that x=$30,052. What are your null and alternative hypothesis?
ANSWER:
Null hypothesis: Average income of American women who work full – time and only have a high school degree is $31,666. That is µ=31,666
Alternative hypothesis: Average income of American women who work full – time and only have a high school degree is different from 31,666. That is µ≠$31,666
Problem 17.3:
a) Null hypothesis: Students on an average study for 15 hours a week. i.e., µ=15 hours
Alternative hypothesis: Students on an average study for more than 15 hours a week i.e.,µ>15 hours
Problem 18.4 :
n=477
Sample mean = 157
Standard deviation = 35
Z for 95% confidence interval is 1.96
95% confidence interval is (Sample mean ± Z(α/2)*Standard error)
Standard error = standard deviation/sqrt(n)
=35/sqrt(477)
= 1.60
Confidence interval = (157±(1.96)*(1.60))
=(153.864,160.136)
c) We are given sample mean and not population mean. Hence we will not able to decide on our trust upon the population mean. If the population mean is given and it falls within the confidence interval, we can place our trust on it with 95% confidence. If not no.
Problem 16.28:
We all know that the Standard error = standard deviation/ sqrt(n). So when the sample size increases, the standard error decreases. Similarly, when sample size decreases, the standard error increases. It is very important to keep the standard error low. This is one of the reasons why Statisticians keep the sample size large.
The margin of error is Z*standard deviation/sqrt(n). In this case also we see that margin of error is inversely proportional to sample size.
Also, practically, we can trust on a particular response when there are repeated experiments or when we have many units are being experimented.
These reasons make the Statistician prefer larger sample sizes.
Effect of sample size on margin of error:
Margin of error = Z(σ/sqrt(n))
Consider Z(α/2) for 95% confidence level is 1.96.
Consider n=8
Margin of error = σ*(1.96/sqrt(8) = 0.6929*σ
For n=50
Margin of error = σ*(1.96/sqrt(50)
=0.2771*σ
From the above we see that when n=8, margin of error is .6929 time the standard deviation
When n=50, margin of error is 0.2771 times the standard deviation. Thus sample size has an inverse effect on the margin of error.
Problem 18.14:
a) Power of the test is nothing but the likelihood of rejecting the null hypothesis when in fact it is false. That is when there is a significant effect the test rightly rejects the null hypothesis.
In this case the TUDA has measured the reading progress of eighth graders as 243. But when a random sample yielded the reading progress as 255, there is a doubt on this. Hence a study was undertaken. Rejecting the null hypothesis means to place trust on the reading progress being greater than 243. Probability of rightly rejecting the null is 0.29 which is the power of the test.
