5.1
(a) the population is the uniform distribution over integers 1 to 5. The mean and variance of the population are and
(b)
(1,1)(1,2)
(1,3)
(1,4)
(1,5)
(2,1)
(2,2)
(2,3)
(2,4)
(2,5)
(3,1)
(3,2)
(3,3)
(3,4)
(3,5) / 1.0
1.5
2.0
2.5
3.0
1.5
2.0
2.5
3.0
3.5
2.0
2.5
3.0
3.5
4.0 / (4,1)
(4,2)
(4,3)
(4,4)
(4,5)
(5,1)
(5,2)
(5,3)
(5,4)
(5,5) / 2.5
3.0
3.5
4.0
4.5
3.0
3.5
4.0
4.5
5.0
(c)
/ 1.0 / 1.5 / 2.0 / 2.5 / 3.0 / 3.5 / 4.0 / 4.5 / 5.0(d) The mean of is
The variance of is
These equal and , respectively.
5.34 Since the rth order statistic in a random sample of size n from a U[0,1] distribution has the beta distribution with parameters r and n-r+1, , , and
5.38
(a)
(1,2)(1,3)
(1,4)
(1,5)
(2,3)
(2,4)
(2,5) / 1.5
2.0
2.5
3.0
2.5
3.0
3.5 / (3,4)
(3,5)
(4,5) / 3.5
4.0
4.5
(b)
/ 1.5 / 2.0 / 2.5 / 3.0 / 3.5 / 4.0 / 4.5(c) The mean of is
The variance of is
(d) The finite population correction factor is . Then the variance of is . This matches the answer in (c).
6.2
(a)
(b)
Therefore has the smallest variance.
(c)
So is unbiased when
is minimum when because subject to , is minimized by choosing .
6.4
6.11
(a) Since , a 90% CI for is
(b) 90% CI
(c) The probability that is included in this CI is either 0 or 1 (not .90).
6.13
(a) You would expect 100X0.95=95 intervals to contain the true mean 70.
(b) X~Bin(100,0.95).
6.16
Conf. Level :
For n=10,
Conf. Level :
6.31
For case (i), and P-value = 0.159
For case (ii), and P-value = 0.023
For case (iii), and P-value = 0.001
As the sample size gets large, the P-value gets small, implying that even small differences from the hypothesized mean will be found to be significant if the sample size is large enough.
7.1
(a)
From equation (7.5)
He needs to catch 56 fish.
(b)
If n=100, then
The new margin of error is or 74% of the old margin of error.
7.4
(a)
Range=$350-$10=$340. Therefore, a rough estimate of . For a 95% CI with a margin of error of 10, using equation (7.5),
Therefore, 278 incorrect orders need to be sampled.
(b)
For a 99% CI with a margin of error of 10,
Therefore, 480 incorrect orders needs to be sampled.
7.5
(a)
The appropriate hypotheses are :
vs
(b)
The observed test statistic is
Then the P-value is
Our conclusion is not to reject at level since the P-value (although it is on the borderline). There is not sufficient evidence that the mean yield for Illinois differs from the national average.
(c)
The assumption that the 50 farmers are a random sample from the population of Illinois farmers is far more important. If they do form a random sample, then the sample mean is approximately normal regardless of the original distribution of the Illinois yields as a consequence of the central limit theorem.
7.6
(a)
The appropriate hypotheses are :
ozs. vs ozs.
The alternative should be two-sided, since the mean could shift in either direction.
(b)
The decision rule is to reject if
i.e. if
i.e. if either
or
(c)
If , the power for this 2-sided test is, from equation(7.9),
P(reject
(d)
To assure 90% power for detecting a difference of 0.1 ozs, use =1-Power=0.1. Then, from equation (7.11),
Therefore, 11 cans should be sampled.
7.7
(a)
The observed test statistic is
Since this is a 1-sided test, the P-value is
Our conclusion is not to reject at level since the P-value. There is not sufficient evidence that the mean tire life for this sample differs from average tire life for the old tread design.
(b)
If , the power for this 1-sided test is, from equation (7.7),
(c)
To assure 90% power in detecting a mean wear of 61000 miles, use =1-Power=0.1. Then, from equation (7.10),
Therefore, 30 tires should be tested.
7.10
(a)
The appropriate hypotheses are :
vs.
(b)
The range is 4 mph. Therefore, a rough estimate of . To assure 95% power for detecting a bias of 0.5 mph or greater, use =1-Power=0.05. Then, from equation (7.11),
Therefore, 72 speedometers should be tested.
(c)
The test statistic is
Since , our conclusion is to not reject at level . There is not sufficient evidence that the speedometers have a bias.
(d)
If the bias is 0.5 mph, the power for this 2-sided test is
7.11
(a) You would expect 95 of the 95% z-intervals to contain the true mean
(b) We still expect 95 of the 95% t-intervals to contain the true mean. The confidence intervals are developed so that 95% of them contain the true mean on average, regardless of the type of interval.
7.13
(a)
Using and s=0.518, a lower 95% confidence bound is given by
Since this lower confidence bound exceeds 87, we would conclude that the mean octane rating exceeds 87.
(b)
The hypotheses are vs. . The test statistic is
Since , the P-value lies between 0.005 and 0.001. Therefore this result would be significant at but not at
7.15
(a) The parameter refers to the true average proportion of students using the food service.
(b)
and s=6.66. The test statistic then is
Since , our conclusion is to reject very strongly at level . There is sufficient evidence that the mean usage of the food service has increased as of the 4소month of the contract.
(c)
The appropriate hypotheses are
vs.
(d)
The test statistic is
Since , our conclusion is not to reject at level . There is not sufficient evidence that the food service has met its goal of at least 70% usage.
7.17
(a) The straight line normal plot indicates that the data follow a normal distribution.
(b)
The test statistic is
Since , we reject the null hypothesis. There is sufficient evidence that the precision of the new device is better than the current monitor.
(c)
An upper one-sided confidence interval for is given by equation (7.20),
Then an upper one-sided confidence bound for is . Since this is less than , our conclusion is to reject . This is consistent with our conclusion from the hypothesis test.