MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
Find the critical value
corresponding to sample size of 5and confidence level of 98
Find the critical value
corresponding to samplesize of 24and confidence level of 95
Use the given degree of confidence and sample data to find confidence interval for the population standard
deviation o. Assume that the population has normal distribution Round the confidence interval limits tothe same
number of decimal places as the sample standard deviation.
College students annual earnings: 98% confidence; n = 9. 3959 886
A) $539 1734
Provide an appropriate response.
4) Suppose that you wish to test claim about apopulation mean. Which distribution should be
used given that the sample isa simple random sample, ois unknown, n = 15, and the population
isnot normally distributed?
A) Neither the normal northe t-distribution
C) Normal distribution
Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol (p, p. the
Anentomologist writes an articlein scientific journal which claims that fewer than 7inten
thousand male fireflies are unable to produce light due toa genetic mutation. Use the parameter
P. the true proportion of fireflies unable to produce light
A) Ho 0.0007
B) Ho 0,0007
C) Ho 0.0007
H1 11:p> 0.0007
6) Carter Motor Company claims that its new sedan the Libra, will average better than 26 miles
per gallon the city Use A, the true average mileage of the Libra.
A) Ho 26
B) Ho 26
7) A researcher claims that the amounts of acetaminophen in certain brand of cold tablets have a 7)
standard deviation different from the mg claimed by themanufacturer.
B) Ho o<3.3 mg
C) Ho O> mg
Assume that the data has normal distribution and the number of observations is greater than fifty. Find the critical z
value used to test null hypothesis.
8) 0.05for two-tailed test
9) 0.09 for right-tailed test.
10) 0.05fora left tailed test.
Find the value of the test statistic using
11) claimi made that the proportion of children who play sportsi less than 0.5, and the sample 11)
statistics include 1671 subjects with 30% saying that they playa sport.
Use the given information to find the value. Also, usea 0.05 significance level and state the conclusion about the
null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).
12) The test statistici right-tailed test is 0.52.
A) 0.3015 reject the null hypothesis
B) 0.0195; reject the null hypothesis
C) 0.6030; fail to reject the null hypothesis
D) 0.3015; fail to reject the null hypothesis
13) The test statistic in aleft-tailed test isz=- .83.
A) 0.9664; fail to reject the null hypothesis
B) .0336; reject thenull hypothesis
C) 0.0672; reject the null hypothesis
D) 0.0672; fail to reject the null hypothesis
14) The test statistic in test is z = 1.95
A) 0.9744; fail to reject the null hypothesis
B) .0256; reject the null hypothesis
C) 0.0512; reject the null hypothesis
D) 0.0512; fail reject the null hypothesis
Assume that hypothesis test of the given claim will be conducted Identify the typel or type II error for the test.
15) medical researcher claims that 6% of children suffer from acertain disorder. Identify the type
I error for the test.
A) Reject the claim that the percentage of thildren who suffer from the disorderi equal to 6%
when that percentagei actually 6%.
B) Fail reject the claim that the percentage of children who suffer from the disorder is equal
to 6% when that percentage is actually 6%.
C) Fail to reject the claim that the percentage of children who suffer from the disorder equal
to 6% when that percentage actually different from 6%.
D) Reject the claim that the percentage of children who suffer from the disorderis different
from 6% when that percentage really different from 6%.
16) The principal of school claims that the percentage of students at his school that come from
single- parent homes is 13% Identify the type error for the test.
A) Reject the claim that the percentage of students that come from single-parent homes is
equa to 13% when that percentage is actually less than 13%
B) Reject the claim that the percentage of students that come from single-parent homes is
equal to 13% when that percentage actually 13%.
C) Fail reject the claim that the percentage of students that come from single parent homes
is equal to 13% when that percentageis actually 13%
D) Fail reject the claim that the percentage of students that come from single parent
is equal to 13% when that percentage actually different from 13%.
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Here we need to use df=n-1=5-1=4, and look it up in the table for p=0.99
Here we need to use df=n-1=23 and look it up in the table for p=0.025
Here we need to use the formula for confidence interval for standard deviation.
Chi_R= 20.0900, Chi_L= 1.6460, obtained from the table, use df=n-1=,p=99 or p=.01...