Question
Mary, a javelin thrower, claims that her average throw is 61 meters. During a
practice session, Mary has a sample throw mean of 55.5 meters based on 12
throws. At the 1% significance level, does the data prov
...
Question
Mary, a javelin thrower, claims that her average throw is 61 meters. During a
practice session, Mary has a sample throw mean of 55.5 meters based on 12
throws. At the 1% significance level, does the data provide sufficient evidence to
conclude that Mary's mean throw is less than 61 meters? Accept or reject the
hypothesis given the sample data below.
H0:μ=61 meters ; Ha:μ<61 meters
α=0.01 (significance level)
z0=−1.99
p=0.0233
Well done! You got it right.
Reject the null hypothesis because |−1.99|>0.01 .
Do not reject the null hypothesis because |−1.99|>0.01 .
Reject the null hypothesis because the p -value 0.0233 is greater than the
significance level α=0.01 .
Do not reject the null hypothesis because the value of z is negative.
Do not reject the null hypothesis because the p -value 0.0233 is greater than the
significance level α=0.01 .
Answer Explanation
Correct answer:
Do not reject the null hypothesis because the p -value 0.0233 is greater than the
significance level α=0.01 .
In making the decision to reject or not reject H0 , if α>p -value, reject H0 because
the results of the sample data are significant. There is sufficient evidence to
conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha ,
may be correct. If α≤p -value, do not reject H0 . The results of the sample data arenot significant, so there is not sufficient evidence to conclude that the alternative
hypothesis, Ha , may be correct. In this case, α=0.01 is less than or equal to
p=0.0233 , so the decision is to not reject the null hypoth
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
80.6840.7190.090.5750.6140.6520.6880.722Well 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, wewant 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
p=0.018
Great work! That's correct.
Do not reject the null hypothesis because 2.1>0.05 .Do not reject the null hypothesis because the value of z is positive.
Reject the null hypothesis because 2.75>0.05 .
Reject the null hypothesis because the p -value 0.018 is less than the significance
level α=0.05 .
Do not reject the null hypothesis because the p -value 0.018 is less than the
significance level α=0.05 .
Answer Explanation
Correct answer:
Reject the null hypothesis because the p -value 0.018 is less than the significance
level α=0.05 .
In making the decision to reject or not reject H0 , if α>p -value, reject H0 because
the results of the sample data are significant. There is sufficient evidence to
conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha ,
may be correct. If α≤p -value, do not reject H0 . The results of the sample data are
not significant, so there is not sufficient evidence to conclude that the alternative
hypothesis, Ha , may be correct. In this case, p=0.018 is less than the
significance level α=0.05 , so the decision is to reject the null hypothesis.
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
Kurtis is a statistician who claims that the average salary of an employee in the city
of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this
to be incorrect, so she randomly selects 61 employees who work in Yarmouth and
record their annual salary. Gina calculates the sample mean income to be
$56,500 per year with a sample standard deviation of 3,750 . Using the
alternative hypothesis Ha:μ>55,000 , find the test statistic t and the p-value for
the appropriate hypothesis test. Round the test statistic to two decimal places and
the p-value to three decimal places.
Right-Tailed T-Table
probability 0.0004 0.0014 0.0024 0.0034 0.0044 0.0054 0.0064
Degrees of Freedom
54 3.562 3.135 2.943 2.816 2.719 2.641 2.576
55 3.558 3.132 2.941 2.814 2.717 2.640 2.574
56 3.554 3.130 2.939 2.812 2.716 2.638 2.572
57 3.550 3.127 2.937 2.810 2.714 2.636 2.571
58 3.547 3.125 2.935 2.808 2.712 2.635 2.569
59 3.544 3.122 2.933 2.806 2.711 2.633 2.568
60 3.540 3.120 2.931 2.805 2.709 2.632 2.567
Great work! That's correct.
t is equal to 3 point 1 2 comma p-value is equal to 0 point 0 0 1$$t=3.12, pvalue=0.001t is equal to 3 point 1 2 comma p-value is equal to 0 point 0 0 1 - correct
Correct answers:
Answer Explanationt is equal to blank comma p-value is equal to$$t= , pvalue= t is equal to blank comma p-value is equal to - no
response given
Correct answers:
0t is equal to 3 point 1 2 comma p-value is equal to 0 point 0 0 1$t=3.12,\
\text{p-value}=0.001$t=3.12, p-value=0.001
Since σ is unknown and the sample size is at least 30 , the hypothesis test for the
mean can be performed using the t -distribution. Here, the sample mean x¯¯¯ is
56,500 , the hypothesized mean μ is 55,000 , the sample standard
deviation s is 3,750 , and the sample size n is 61 . Substitute these values into
the formula to calculate the t test statistic.
tt=56,500−55,0003,750/61−−√≈3.12
Now find the p-value. Notice that the test statistic has 61−1=60 degrees of
freedom and that this is a right-tailed test because the alternative hypothesis is
Ha:μ>55,000 . Find the p-value for a right-tailed test of a t -distribution with
60 degrees of freedom, where t≈3.12 . That is, to find the p-value, find the area
under the t -distribution curve with 60 degrees of freedom to the right of t≈3.12
. The p-value that corresponds to these conditions is approximately 0.001 .
Question
What is the p -value of a two-tailed one-mean hypothesis test, with a test statistic
of z0=−1.59 ? (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
.010.0350.0440.0540.0
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