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STAT 200
Guided Exercise 6
1. The EPA standard on the amount of suspended solids that can be discharged into a river is a
maximum of 60mg per liter per day (mg/L). You want to test a randomly selected sample of n water
specimens and estimate the mean daily rate of pollution produced by a mining operation. You have been
asked to submit a proposal to do this study. The granting agency wants to know about the precision of
your estimate Suppose you want a 95% C.I. with a bound of error (B) of .8 mg/l.
The formula for solving for n is the following. Please note that it is an algebraic
manipulation of the confidence interval formula to solve for n based on a desired plus
or minus bound of error of .8 mg/l.
a. What is the minimum sample size you would need? Assume the water readings are approximately
normal with σ = 5 mg/L
α/2 = .025 since (1-.95) =.05; z= 1.96; B = .8 mg/l
N = 151 as a minimum sample size
b. Check you answer by calculating the confidence interval for this problem (the Bound of Error) using the
sample size you calculated.
2. A larger supermarket began to get complaints from customers about the quantity of chips in a 16ounce bag of a particular brand of potato chips. The chain decided to test the following hypothesis
concerning µ, the mean weight (in ounces) of a bag of potato chips in the next shipment of chips received
from their supplier.
Ho: µ = 16
Ha: µ < 16
If there is evidence that µ < 16, then the shipment would be refused and a complaint would be registered
with the supplier.
a. What is the Type I error, expressed in terms of this problem.
Refusing the shipment when the weight of the chips was ok.
b. What is Type II error, expressed in terms of this problem.
Accepting the shipment when in fact the chips were underweight.
b. Which type of error would the customers view as more serious?
The customers would rather err on the side of being cautious and reject more shipmentsThey would be concerned with Type II Error.
d. Which type of error would the supplier view as more serious?
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The supplier would rather have their product accepted –
They would be concerned with Type I Error, rejecting shipments when the weight was ok.
3. Current technology uses X-rays and lasers for inspection of solder-joints on printed circuit boards.
A current manufacturer claims its product can inspect on average at least 10 solder joints per second. A
potential buyer tested the equipment on 27 different printed circuit boards (PCB). The sample mean and
standard deviation are given below. We will assume that this distribution is approximately normal.
This is a small sample hypothesis test of a mean.
Mean = 10.54
s = 1.61
n = 27
a. Calculate a 95% confidence interval for this sample estimate.
Standard Error = 1.61/ (27).5 = .3098; t.05/2, 26 d.f. = 2.056
10.54 ± 2.056(.3098)
10.54 ± .637
9.903 to 11.177
b. Test to see if the sample data is different from the manufacturer’s claim. You will need to determine
the Null Hypothesis and your Alternative Hypothesis. Use α = .05. Note: use the t-distribution for
the rejection region.
HYPOTHESIS TEST
Null Hypothesis
H0 : µ=10
Alternative Hypothesis
Ha : µ≠10 Two-tailed Test
Assumptions of Test
n=27 is a small sample. So I use the t-distribution, with α=0.05/2
and 27-1=26 d.f. We assume the distribution in the population is
approximately normal.
Test Statistic (z* or t*)
t*=(10.54-10.0)/.3098
Rejection Region
t.05/2,26 d.f.=2.056
Calculation of Test Statistic
t*=1.743
Comparison of Test
Statistics with Rejection
Region
-2.056<t*<2.056
Cannot reject H0
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4. A large city’s Department of Motor Vehicles (DMV) claimed that 80% of candidates pass driving
tests on the first try. A newspaper surveyed 90 randomly selected local teens who had taken the test
and found 61 teens had passed on the first try. This value seemed lower than the one reported by DMV.
Test to see if the passing rate for teenagers from our sample is less than the overall rate.
a. What is the proportion for our sample?
Est. p = 61/90 = .6778
b. Determine if we can use the normal approximation of the binomial. Remember, n*p or n*q should be >
5 in order to use this approach.
90*.6778 = 61
90*.3222 = 29
c. If the Null Hypothesis is .80, what is the standard error for this problem?
Use the value under Ho: p = .80
S.E. = SQRT[(.80*.20)/90] = SQRT[.0018] = .0422
d. Now finish the steps for the hypothesis test. Use an alpha level of .05.
HYPOTHESIS TEST
Null Hypothesis
p = .80
Alternative Hypothesis
p < .80
Assumptions of Test
use normal approximation to binomial
Test Statistic (z* or t*)
z*= (.6778 - .80)/.0422
Rejection Region
z.05 = -1.645
Calculation of Test Statistic
z* = -2.90
Comparison of Test
Statistics with Rejection
Region
z* < z.05 ; Reject H0: p = .80 ; We have evidence that the
sample proportion is lower than .80.
e.
Take the test statistic that you calculated (z*) and find the probability associated with this test statistic
value in the standard normal table. The probability of the test statistic is thought of the probability of
that value and further out in the tail. For this problem, it is .5 - p(ABS[z*]). We use the absolute value
since our z-score is negative. This value is known as a p-value for this problem.
We want to compare the p-value to the alpha level set for the problem. If it is lower than alpha, we can reject
the Null Hypothesis. Does the p-value approach match your conclusion above?
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With a z* = -2.90 the p-value is:
.5 - .4981 = .0019 This is much lower than α = .05, so we can reject the Null Hypothesis
What if our alternative hypothesis was a two-tailed test, what would the p-value be?
If this is a two-tailed test, p= .0019*2 = .0038
5. Therapeutic touch, taught in many schools of nursing, is a therapy in which the practitioner moves
her hand near, but does not touch, a patient in an attempt to manipulate a “human energy field.”
Therapeutic touch practitioners believe that by adjusting this field they can promote healing. A study was
done to see if there is any evidence of Therapeutic touch. The experiment involved 15 TT practitioners
who were asked to determine if they could detect an unseen hand hovering over their right or left hand.
Practice sessions were used at first where the practitioners could see the hovering hand. Then a screen
was placed over the hands and 10 trials were conducted for each practitioner. A coin flip was used to
randomly determine which hand to hover over. Overall, 150 trials were conducted and the TT practitioners
were successful 70 times, for a proportion of .467.
a. Is there evidence from this experiment that TT practitioners can successfully detect a “human energy
field?” You need to conduct a hypothesis test for this proportion problem. Use an alpha level of .01.
HYPOTHESIS TEST
Null Hypothesis
p = .50
Alternative Hypothesis
p >.50
Assumptions of Test
use normal approximation to binomial
Test Statistic (z* or t*)
z*= (.467 - .50)/SQRT[.5*.5/150]
Rejection Region
z.01 = 2.33
Calculation of Test Statistic
z* = -.8083
Comparison of Test
Statistics with Rejection
Region
z* < z.01
Cannot Reject Ho: p = .50
We do not have evidence that the sample proportion is any
different from .50.
b. Calculate the p-value for this test.
This one is a little tricky, since we expected the est. p to be above .5 when in fact it was below .5
different
from .50.
(The TT practitioners did worse than
chance).
So the p-value is bases on z* = -.81
Lookup in table .2910 Plus .5 = .7910
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6. Speeders on the highway. The following is data from a random sample of 63 motorists on a Federal
highway where the speed limit of 65 mph. The data can be found on an Excel file on the web
(speeders.xls).
a. Below is the Excel output from the descriptive statistics of the speed of drivers and the Stem and Leaf
Plot from JMP. Briefly (one paragraph) describe the data using the summary statistics.
The mean speed for the 63 motorists was 69.2 mph with a range of 39. The highest speed was 95 and
the lowest was 56.
The distribution is approximately mound shaped and symmetrical and the mean, median, and mode
are all very similar, however the mean is slightly higher than the median and is being pulled by a few
extreme high speeds.
The Stem and Leaf plot shows the distribution is skewed right toward some high values.
There is a fair amount of variability in the data with a standard deviation of 8.
b. Conduct a Hypothesis Test where the null hypothesis is that the average speed is 65 mph. You are free to
decide if it should be a one or two tailed test. Use an α level of .01.
HYPOTHESIS TEST
Null Hypothesis
µ = 65
Alternative Hypothesis
µ > 65
Assumptions of Test
N is large (63) so we can use a z-value
Test Statistic (z* or t*)
(69.22-65)/1.016
Rejection Region
Z= 2.33
Calculation of Test Statistic
Z* = 4.1535
Comparison of Test
Statistics with Rejection
Region
Z* > z We can reject Ho: µ = 65; There is evidence to suggest
people are speeding
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7. Golf course designers are worried that the new equipment is making old courses obsolete. One
designer says that courses need to be built with the expectation that players will be able to drive the ball an
average of 250 yards or more. A sample of 135 golfers is taken and they measured their driving distance:
n=135
mean = 256.3 yards
s = 43.4 yards
a. Does the sample provide enough evidence to suggest that golfers are already hitting it farther than the
250 mark? Use α=.05
HYPOTHESIS TEST
Null Hypothesis
H0:µ= 250
Alternative Hypothesis
Ha: µ > 250 one-tailed test, upper
Assumptions of Test
Large sample, normal
Test Statistic (z* or t*)
z* = (256.3-250)/(43.4/135.5)
Rejection Region
z =.05 = 1.645
Calculation of Test Statistic
z* = 1.69
Comparison of Test
Statistics with Rejection
Region
z* > za=.05
1.69 > 1.645
Reject H0: µ= 250
α
b. Take the test statistic that you calculated (i.e., t* or z*) and find the probability associated with this test
statistic value. If it is a z-value, the probability will be:
.5 - p(z*)
this is known as a p-value for this problem
z* = 1.69
P(0<x<z*) = .4545
p= .5 - .4545 = .0455
Compare this to the alpha level set for the problem.
α =.05
The p-value is less than .05, therefore I have a z* that is out in the rejection region
When the p-value is less than α, I can reject the Null Hypothesis
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