Unformatted text preview: CE93  Engineering Data Analysis Prof. Joan Walker, Spring 2011 Solutions to Assignment 10
1. Average speed of vehicles on a highway
a. Observations, n = 50.
confidence interval for the mean speed, . . Determine twosided 99% Because n is large, the distribution of
For 99% confidence interval, Confidence interval for b. What value of n gives a 99% confidence interval such that we know
within ±1. That is, what n gives a confidence interval (65 – 1 , 65 + 1) =
(64, 66). However, n must be integer, so we take the first integer value larger than
the number found above. Therefore, n = 240. The additional vehicles
observed is n – 50 = 240 – 50 = 190.
c. For n=n1=n2=100, the respective sample means are normally distributed: .
The difference in sample means are distributed normal as well: CE93  Engineering Data Analysis Prof. Joan Walker, Spring 2011 Where 2. 30 observations (n = 30), mean = 22 min (
, sample variance = 4.2
2
min (
We want a twosided 99% confidence interval for the
mean,
For a sample with large n, the sample mean is normally distributed. When the
population variance is unknown, the following statistic is follows a tdistribution
with n1 degrees of freedom: Confidence interval for : CE93  Engineering Data Analysis Prof. Joan Walker, Spring 2011 3. Bulb lifetime – population is normally distributed.
a. An estimate for the population variance is the sample variance, b. Twosided, 99% confidence interval for
For 99% confidence interval, : . The following statistic follows a chisquare distribution with n1 degrees
of freedom: Twosided confidence interval for : Confidence interval for c. For 90% confidence that 4. Binomial trails with successes 6, 5, 9, and 7. Let’s denote the number of
experiments as N = 4. The set of independent trails represents a binomial
experiment with trials n and success probability p. CE93  Engineering Data Analysis Prof. Joan Walker, Spring 2011 a. For n = 12, determine the maximum likelihood for the success probability,
p. For a binomial random variable with n trails, and success probability p,
its probability mass function (for x successes) is given by: The likelihood function is simply the product of the pmf for each separate set of
trials. Taking the logarithm gives us the log likelihood: The log likelihood is differentiable with respect to p, so let’s do that and set it
equal to zero. Solving for p: For this problem,
For n = 12, we get the result that
To ensure that this is a maximum, we can check the second order condition.
Taking another derivative with respect to p (Not required to get full credit): CE93  Engineering Data Analysis Plugging in , we get Prof. Joan Walker, Spring 2011 , so our estimate for is a maximum. b. In part (a) we found the maximum likelihood estimate to be
.
However, our choice of probability is now constrained to be either 0.5 or
0.75. To determine which probability is more likely, we can calculate the
likelihood (or log likelihood) of each one, and see which is larger. The
probability that produces a larger likelihood is the maximum likelihood
estimate. For p = 0.5, For p = 0.75, The likelihood for p = 0.6 is larger, so our maximum likelihood estimate is c. Now we know that the true probability of success is p = 0.8. We need to
determine a maximum likelihood estimate for the number of trials, n. The
likelihood function is the same as before This likelihood is not differentiable with respect to n, however. The combinations
term includes factorials which are not easily differentiable. But we know certain
constraints for n that will help us find a solution. First of all, n must be a positive
integer, because it is representing the number of discrete trials. Secondly, n must
be at least as large as the largest number of successes found. In this case, the
largest number of successes is 9. CE93  Engineering Data Analysis Prof. Joan Walker, Spring 2011 We will start out with n = 9, and increase n until we find a maximum value of the
likelihood function.
For n = 9: For n = 10: For n = 11: The likelihood is decreasing as we increase n. The largest value is found when n =
9, so our maximum likelihood estimate for n is 5. Part a: Upper 1sided 95% confidence interval: Lower 1sided 95% confidence interval: CE93  Engineering Data Analysis Prof. Joan Walker, Spring 2011 Part b: 2sided 95% confidence interval: 6. Twosided hypothesis test: H0:µ = 8.2 vs. H1:µ ≠ 8.2 3.3204 (use the normal distribution because variance is
known)
pvalue = 2 [1 − φ(3.3204)] = .0010
This is greater than both α=0.1 and α=0.05, so we can reject H0 at both levels of
significance.
7. Twosided hypothesis test:
H0:µ = 30 vs. H1:µ ≠ 30 1.847 (use the tdistribution because variance is unknown)
pvalue = P{T15 < −1.847} = .04
So we cannot reject H0 at a 95% level of significance. Note: We do not
necessarily have to perform a twosided test. If we conducted onesided ttest at a
95% level of significance, we would reject H0.
In this case, we only care about the sheets not meeting the stress resistance
specification, but we don’t really care if they exceed the standard. In addition,
, so it we have reason to suspect that the sheets do not meet the
specification. In this case, it would be justified to use a onesided test, which has
the stricter rejection criterion. CE93  Engineering Data Analysis Prof. Joan Walker, Spring 2011 8. Twosided hypothesis test:
Denote X: pH level of Solution A; Y: pH level of Solution B.
X
and Y
n=10, m=10,
and
Twosample problems with known variances
1 twosided confidence interval for is 5% level of significance: 95% confidence interval is
= Define a random variable
With , and 95% twosided confidence interval for is The hypothesis test is set up as
vs
which can also be treated as
vs (a) Method 1:
is in the 95% confidence interval of
Method 2: , we do not reject H0. CE93  Engineering Data Analysis Prof. Joan Walker, Spring 2011 So we fail to reject H0, and we conclude that the data support the claim.
(b) p_value = 2*p(Z<TS)=2*
0.4209
(p_value is the smallest level that we can reject H0. Since p_value is larger than
the significance level 0.05, we fail to reject H0.) ...
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This note was uploaded on 06/02/2011 for the course CIV ENG 93 taught by Professor Staff during the Spring '08 term at Berkeley.
 Spring '08
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