Construct Hypothesis Test for
Proportions
Question
What is the p -value of a two-tailed one-mean hypothesis test, with a test statistic
of z0=−1.73 ? (Do not round your answer; compute your answer using a value
from the table below.)
z−1.8−1.7−1.6−1.5−1.40.000.0360.0450.0550.0670.0810
.010.0350.0440.0540.0660.0790.020.0340.0430.0530.0640.
0780.030.0340.0420.0520.0630.0760.040.0330.0410.0510.
0620.0750.050.0320.0400.0490.0610.0740.060.0310.0390.
0480.0590.0720.070.0310.0380.0470.0580.0710.080.0300.
0380.0460.0570.0690.090.0290.0370.0460.0560.068
Well done! You got it right.
0 point 0 8 4$$0.0840 point 0 8 4 - correct
Correct answers:
Answer Explanation
$$no response given
Correct answers:
, 00 point 0 8 4$0.084$0.084
The p -value is the probability of an observed value of z=1.73 or greater in
magnitude if the null hypothesis is true, because this hypothesis test is two-tailed.
This means that the p -value could be less than z=−1.73 , or greater than
z=1.73 . This probability is equal to the area under the Standard Normal curve
that lies either to the left of z=−1.73 , or to the right of z=1.73 .
A normal curve is over a horizontal axis and is centered on 0. Two points are
labeled negative 1.73 and 1.72 The area to the right of 1.73 and to the left of
negative 1.73 is shaded.
Using the Standard Normal Table given, we can see that the p -value that
corresponds with z=−1.73 is 0.042 , which is just the area to the left of
z=−1.73 . Since the Standard Normal curve is symmetric, the area to the right of
z=1.73 is 0.042 as well. So, the p -value of this two-tailed one-mean hypothesis
test is (2)(0.042)=0.084 .
Question
What is the p -value of a two-tailed one-mean hypothesis test, with a test statistic
of z0=0.27 ? (Do not round your answer; compute your answer using a value
from the table below.)
,z0.10.20.30.40.50.000.5400.5790.6180.6550.6910.010.54
40.5830.6220.6590.6950.020.5480.5870.6260.6630.6980.0
30.5520.5910.6290.6660.7020.040.5560.5950.6330.6700.70
50.050.5600.5990.6370.6740.7090.060.5640.6030.6410.67
70.7120.070.5670.6060.6440.6810.7160.080.5710.6100.64
80.6840.7190.090.5750.6140.6520.6880.722
Well done! You got it right.
0 point 7 8 8$$0.7880 point 7 8 8 - correct
Correct answers:
Answer Explanation
$$no response given
Correct answers:
00 point 7 8 8$0.788$0.788
The p -value is the probability of an observed value of z=0.27 or greater in
magnitude if the null hypothesis is true, because this hypothesis test is two-tailed.
This means that the p -value could be less than z=−0.27 , or greater than
z=0.27 . This probability is equal to the area under the Standard Normal curve
that lies either to the left of z=−0.27 , or to the right of z=0.27 .
, A standard normal curve with three points labeled on the horizontal axis labeled z.
The mean is labeled at 0 and observed values of negative 0.27 and 0.27 are labeled.
The areas under the curve and to the left of negative 0.27 and to the right of 0.27
are shaded. The shaded areas are both labeled p-value.
Using the Standard Normal Table, we can see that the p -value that corresponds
with z=0.27 is 0.606 , which is the area to the left of z=0.27 . However, we
want the area to the right of 0.27 , which is 1−0.606=0.394 . Because the
Standard Normal curve is symmetric, the area to the left of z=−0.27 is 0.394
as well. So, the p -value of this two-tailed one-mean hypothesis test is (2)
(0.394)=0.788 .
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Question
Marty, a typist, claims that his average typing speed is 72 words per minute.
During a practice session, Marty has a sample typing speed mean of 84 words per
minute based on 12 trials. At the 5% significance level, does the data provide
sufficient evidence to conclude that his mean typing speed is greater than 72
words per minute? Accept or reject the hypothesis given the sample data below.
H0:μ≤72 words per minute ; Ha:μ>72 words per minute
α=0.05 (significance level)
z0=2.1
Proportions
Question
What is the p -value of a two-tailed one-mean hypothesis test, with a test statistic
of z0=−1.73 ? (Do not round your answer; compute your answer using a value
from the table below.)
z−1.8−1.7−1.6−1.5−1.40.000.0360.0450.0550.0670.0810
.010.0350.0440.0540.0660.0790.020.0340.0430.0530.0640.
0780.030.0340.0420.0520.0630.0760.040.0330.0410.0510.
0620.0750.050.0320.0400.0490.0610.0740.060.0310.0390.
0480.0590.0720.070.0310.0380.0470.0580.0710.080.0300.
0380.0460.0570.0690.090.0290.0370.0460.0560.068
Well done! You got it right.
0 point 0 8 4$$0.0840 point 0 8 4 - correct
Correct answers:
Answer Explanation
$$no response given
Correct answers:
, 00 point 0 8 4$0.084$0.084
The p -value is the probability of an observed value of z=1.73 or greater in
magnitude if the null hypothesis is true, because this hypothesis test is two-tailed.
This means that the p -value could be less than z=−1.73 , or greater than
z=1.73 . This probability is equal to the area under the Standard Normal curve
that lies either to the left of z=−1.73 , or to the right of z=1.73 .
A normal curve is over a horizontal axis and is centered on 0. Two points are
labeled negative 1.73 and 1.72 The area to the right of 1.73 and to the left of
negative 1.73 is shaded.
Using the Standard Normal Table given, we can see that the p -value that
corresponds with z=−1.73 is 0.042 , which is just the area to the left of
z=−1.73 . Since the Standard Normal curve is symmetric, the area to the right of
z=1.73 is 0.042 as well. So, the p -value of this two-tailed one-mean hypothesis
test is (2)(0.042)=0.084 .
Question
What is the p -value of a two-tailed one-mean hypothesis test, with a test statistic
of z0=0.27 ? (Do not round your answer; compute your answer using a value
from the table below.)
,z0.10.20.30.40.50.000.5400.5790.6180.6550.6910.010.54
40.5830.6220.6590.6950.020.5480.5870.6260.6630.6980.0
30.5520.5910.6290.6660.7020.040.5560.5950.6330.6700.70
50.050.5600.5990.6370.6740.7090.060.5640.6030.6410.67
70.7120.070.5670.6060.6440.6810.7160.080.5710.6100.64
80.6840.7190.090.5750.6140.6520.6880.722
Well done! You got it right.
0 point 7 8 8$$0.7880 point 7 8 8 - correct
Correct answers:
Answer Explanation
$$no response given
Correct answers:
00 point 7 8 8$0.788$0.788
The p -value is the probability of an observed value of z=0.27 or greater in
magnitude if the null hypothesis is true, because this hypothesis test is two-tailed.
This means that the p -value could be less than z=−0.27 , or greater than
z=0.27 . This probability is equal to the area under the Standard Normal curve
that lies either to the left of z=−0.27 , or to the right of z=0.27 .
, A standard normal curve with three points labeled on the horizontal axis labeled z.
The mean is labeled at 0 and observed values of negative 0.27 and 0.27 are labeled.
The areas under the curve and to the left of negative 0.27 and to the right of 0.27
are shaded. The shaded areas are both labeled p-value.
Using the Standard Normal Table, we can see that the p -value that corresponds
with z=0.27 is 0.606 , which is the area to the left of z=0.27 . However, we
want the area to the right of 0.27 , which is 1−0.606=0.394 . Because the
Standard Normal curve is symmetric, the area to the left of z=−0.27 is 0.394
as well. So, the p -value of this two-tailed one-mean hypothesis test is (2)
(0.394)=0.788 .
Something's not right...
There is an error in the instruction or question.
This looks broken...
A graph/image/equation/video isn't working
I cannot enter my answer.
A problem is preventing me from entering an answer to this question.
I have an idea!
I have some feedback/suggestions.
Question
Marty, a typist, claims that his average typing speed is 72 words per minute.
During a practice session, Marty has a sample typing speed mean of 84 words per
minute based on 12 trials. At the 5% significance level, does the data provide
sufficient evidence to conclude that his mean typing speed is greater than 72
words per minute? Accept or reject the hypothesis given the sample data below.
H0:μ≤72 words per minute ; Ha:μ>72 words per minute
α=0.05 (significance level)
z0=2.1
