Self- assessment exam D - Inferential Statistics.-answers
Name: _____________________________________________________
Problem 1.When Mendel conducted his famous genetics experiments with peas, one sample of
offspring consisted of 428 green peas and 152 yellow peas. Find a 90% confidence interval
estimate of the percentage yellow peas.
a) Give the following information if they can be obtained from the information of the problem.
n  __ 580 _____ peas ,   ____ peas, x  ____ peas, x  _ 152 _____ yellow peas
152
 0.26,   0.10 ___
580
b) Write the formula appropriate for constructing the required confidence interval and substitute
the value in the formula.
s  ___ peas ,   ____ peas, p  _______, p̂ 
p̂  z  / 2
answer:
p̂q̂
(0.26)( 0.74)
 0.26  1.645
 0.26  1.645(0.018)
n
580
________
 0.260  0.02996
 (0.23, 0.29)
c) We conclude with a 90% of confidence that the true percentage of yellow peas is between
____23 % and _29 % . The error of estimation is __________ 3 %
Problem 2 – An economist wants to estimate the true mean income for the first year of work for
college graduates who have had the profound wisdom to take a statistics course. A sample of 400
such incomes shows a mean of $35000 with a standard deviation of $6000. Construct a 99%
confidence interval for the true mean income for the first year of work of college graduates who
have had the profound wisdom to take a statistics course.
a) Give the following information if they can be obtained from the information of the problem.
n  400 __,   __________ x  $35000 __, s  $6000 _,   ________,   0.01 _____
b) Write the formula appropriate for constructing the required confidence interval and substitute
the value in the formula.
s
6000
x  z / 2
 35000  2.5758
 35000  2.5758(300)  35000  2.5758(300)
n
400
 35000  773  (34227, 35773)
c) We conclude with a 99% of confidence that the true mean income of college graduates is
between $34,227___ and $35773_ . The error of estimation is $____________
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Problem 3 – An economist wants to estimate the true mean income for the first year of work for
college graduates who have had the profound wisdom to take a statistic course. How many such
incomes must be found if we want to be 99% confident that the sample mean is within $500 of
the true population mean? Assume that a previous study has revealed that for such incomes
  $6000.
a) Give the following information if they can be obtained from the information of the problem.
n  _______,   $6000 ___,   0.01 ____, Error of estimation  $500 ______
b) Write the formula appropriate for determining how many incomes must be foun.
2
2

z
 2.5758  6000 
Answer: n    / 2   
  955.425 ___________________
500


 E 
answer: n = 956______________________________________________
c) 956 incomes must be found if we want to be 99% confident that the sample mean is within
$500 of the true population mean?
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Problem 4: When Gregor Mendel conducted his famous hybridization experiments with peas,
one such experiment resulted in offspring consisting of 428 peas with green pods and 152 with
yellow pods. According to Mendel’s theory, ¼=0.25=25% of the offspring peas should have
yellow pods. Use a 0.05 significance level with the p-value method to test Mendel’s claim that
the proportion of peas with yellow pods is equal to ¼.
Step 1. Is this a hypothesis about p or  ? __p______
n  580 ___,   ___ x  ___, s  ___,   ___, x  152 ___, p̂ 
152
 0.26 _,   0.05 __
580
H 0 : p  0.25 _____ H1 : p  0.25 ________
Step 2. Type of test – select: __a)___________
a) two-tailed test
b) right-tailed test
c) left –tailed test ____________
Step 3.   0.05 _______
Step 4. Find the p-value – Show the work.

p  value  2normalcdf  p̂, 10 9 , p,


  2normalcdf


 2(0.251)
p  value  0.502
pq
n

 0.26, 10 9 , 0.25,


0.25  0.75 

580

p –value = 0.502__
Step 5. Compare the p-value with  . Select: _____b___
a) p-value < 
b) p-value > 
Step 6. Conclusion. Select: __b_______________
a) We reject the null hypothesis
b) We fail to reject the null hypothesis
Select: __a_______________
a) There is not sufficient evidence to warrant rejection at a 0.05 level of the claim that ¼ of the
offspring peas have yellow pods.
b) We have statistical evidence to reject Mendel’s claim that ¼ of the offspring peas should have
yellow pods.
c) Mendel’s claim that ¼ of the offspring peas should have yellow pods is correct
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Problem 5. A Tobacco Company advertised that its best selling non-filtered cigarettes contain at
most 40 mg of nicotine, but Consumer Advocate magazine ran 10 randomly selected cigarettes
and found the amounts (in mg) shown in the accompanying list.
x
47.3
39.3
40.3
38.3
46.3
43.3
42.3
49.3
40.3
46.3
It’s a serious matter to charge that the company advertising is wrong, so the magazine editor
chooses a significance level of   0.01 in testing her believe that the mean nicotine content is
greater than 40 mg. Using a 0.01 significance level, test the editor’s believe that the mean is
greater than 40mg.
a) Find the mean and standard deviation of the given sample (you can use a calculator)
mean = 43.3 mg, standard deviation = 3.80
a) Complete if possible (you can use a calculator)
n  10 ______,   _____ x  43.3 __, s  3.80 ____,   ________,   0.01 ___
Step 1. Is this a hypothesis about p or  ? __ about  ______
H 0 :   40 ______ H1 :   40 _________
Step 2. Type of test – select: Right-tailed test________
a) two-tailed test
b) right-tailed test
c) left –tailed test ____________
Step 3.   0.01 _________
Step 4. Find the p-value – Show the work.
t
x   43.3  40

 2.746
s
3.80
n
10
p  value  tcdf (t , 10 9 , n  1)  tcdf ( 2.746, 10 9 , 9)  0.0113
p –value = 0.0113________
Step 5. Compare the p-value with  . Select:
a) p-value < 
b) p-value > 
b________
Step 6. Conclusion. Select: _____b__________________
a) We reject the null hypothesis
b) We fail to reject the null hypothesis
Select: __b__________________
a) Statistically at a 0.01 level of significance we have evidence to agree with the editor’s believe
that the mean is greater than 40 mg.
b) ) Statistically at a 0.01 level of significance we have evidence to agree with the Tobacco
Company advertised that its best selling non-filtered cigarettes contain at most 40 mg of
nicotine.
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