Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

MATH 225N Week 7 Assignment Conducting a Hypothesis Test for
Mean – Population Standard Deviation Known P-Value Approach
Question
What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1
...
)
z−1
...
7−1
...
5−1
...

0
...
084
The p-value is the probability of an observed value of z=1
...
This means that the p-value could be
less than z=−1
...
73
...
73, or to the right of z=1
...


A normal curve is over a horizontal axis and is centered on 0
...
73 and 1
...
73 and to the left of negative 1
...

Using the Standard Normal Table given, we can see that the p-value that corresponds
with z=−1
...
042, which is just the area to the left of z=−1
...
Since the Standard Normal
curve is symmetric, the area to the right of z=1
...
042 as well
...
042)=0
...
During a practice session,
Mary has a sample throw mean of 55
...
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
...
01 (significance level)
z0=−1
...
0233
Great work! That's correct
...
99|>0
...

Do not reject the null hypothesis because |−1
...
01
...
0233 is greater than the significance
level α=0
...

Do not reject the null hypothesis because the value of z is negative
...
0233 is greater than the
significance level α=0
...

Answer Explanation

Correct answer:
Do not reject the null hypothesis because the p-value 0
...
01
...
There is sufficient evidence to conclude
that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct
...
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
...
01 is less than or equal to p=0
...

Make a conclusion and interpret the results of a one-mean hypothesis test (population
standard deviation known) using the P-Value Approach
Question
Marty, a typist, claims that his average typing speed is 72 words per minute
...
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
...
05 (significance level)
• z0=2
...
018

Select the correct answer below:

Do not reject the null hypothesis because 2
...
05
...

Reject the null hypothesis because 2
...
05
...
018 is less than the significance
level α=0
...

Do not reject the null hypothesis because the p-value 0
...
05
...
Your hard work is paying off 😀😀
Determine the p-value for a hypothesis test for the mean (population standard deviation
known)
Question
What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=0
...
)
z0
...
20
...
40
...

Well done! You got it right
...
788
The p-value is the probability of an observed value of z=0
...
This means that the p-value could be
less than z=−0
...
27
...
27, or to the right of z=0
...


A standard normal curve with three points labeled on the horizontal axis labeled z
...
27 and 0
...
The areas under the curve
and to the left of negative 0
...
27 are shaded
...


Using the Standard Normal Table, we can see that the p-value that corresponds
with z=0
...
606, which is the area to the left of z=0
...
However, we want the area to the
right of 0
...
606=0
...
Because the Standard Normal curve is symmetric, the
area to the left of z=−0
...
394 as well
...
394)=0
...

Determine the p-value for a hypothesis test for the mean (population standard deviation
known)
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
calculates the sample mean income to be $56,500 per year with a sample standard deviation
of 3,750
...
Round the test statistic to two decimal places and the p-value
to three decimal places
...
0004 0
...
0024 0
...
0044 0
...
0064
3
...
558
3
...
550
3
...
544
3
...
135
3
...
130
3
...
125
3
...
120

2
...
941
2
...
937
2
...
933
2
...
816
2
...
812
2
...
808
2
...
805

2
...
717
2
...
714
2
...
711
2
...
641
2
...
638
2
...
635
2
...
632

2
...
574
2
...
571
2
...
568
2
...
Keep it up!
t=3
...
001
Answer Explanation
Correct answers:
• t=3
...
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
...

Substitute these values into the formula to calculate the t test statistic
...
12
Now find the p-value
...
Find the p-value for a
right-tailed test of a t-distribution with 60 degrees of freedom, where t≈3
...
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
...
The p-value that corresponds to these conditions is approximately 0
...

Determine the p-value for a hypothesis test for the mean (population standard deviation
known)
Question
What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1
...
)
z−1
...
7−1
...
5−1
...

Correct! You nailed it
...
112
Answer Explanation
Correct answers:
• 0
...
59 or greater in magnitude if the null
hypothesis is true, because this hypothesis test is two-tailed
...
59, or greater than z=1
...
This probability is equal to the area under the Standard
Normal curve that lies either to the left of z=−1
...
59
...
00 with ticks at negative 1
...
59
...
59 and to the right of 1
...

Using the Standard Normal Table given, we can see that the p-value that corresponds
with z=−1
...
056, which is just the area to the left of z=−1
...
Since the Standard Normal
curve is symmetric, the area to the right of z=1
...
056 as well
...
056)=0
...

Make a conclusion and interpret the results of a one-mean hypothesis test (population
standard deviation known) using the P-Value Approach
Question
Nancy, a golfer, claims that her average driving distance is 253 yards
...
6 yards based on 18 drives
...

• H0:μ=253 yards; Ha:μ<253
• α=0
...
75
• p=0
...
Your hard work is paying off 😀😀
Correct answer:
Do not reject the null hypothesis because the p-value 0
...
02
...
There is sufficient evidence to conclude that H0 is an incorrect
belief and that the alternative hypothesis, Ha, may be correct
...
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
...
02 is less than or equal
to p=0
...

Make a conclusion and interpret the results of a one-mean hypothesis test (population
standard deviation known) using the P-Value Approach
Question
Kathryn, a golfer, has a sample driving distance mean of 187
...
Kathryn
still claims that her average driving distance is 207 yards, and the low average can be attributed
to chance
...


• H0:μ=207 yards; Ha:μ<207
• α=0
...
46
• p=0
...

Correct answer:
Do not reject the null hypothesis because the p-value 0
...
01
...
There is sufficient evidence to conclude that H0 is an incorrect
belief and that the alternative hypothesis, Ha, may be correct
...
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
...
01 is less than or equal
to p=0
...

Make a conclusion and interpret the results of a one-mean hypothesis test (population
standard deviation known) using the P-Value Approach
Question
Ruby, a bowler, has a sample game score mean of 125
...
Ruby still claims that
her average game score is 140, and the low average can be attributed to chance
...

•
•
•
•

H0:μ=140; Ha:μ<140
α=0
...
52
p=0
...

Do not reject the null hypothesis because |−0
...
05
...
3015 is greater than the significance
level α=0
...

Do not reject the null hypothesis because the p-value 0
...
05
...
52|>0
...


Do not reject the null hypothesis because the p-value 0
...
05
Explanation:
Decision rule using the P−value and significance level
...
If P−value>α: Fail to reject H0
...
We compare the P−value with the significance
level (α)
...
3015000000)>α (0
...

Great work! That's correct
...
3015 is greater than the significance
level α=0
...

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
...
If α≤p-value, do not reject H0
...
In this case, α=0
...
3015, so the decision is to not reject the null hypothesis
...
2 from 24 games
...
At
the 5% significance level, does the data provide sufficient evidence to conclude that Marie's
mean game score is less than 143? Given the sample data below, accept or reject the hypothesis
...
05 (significance level)
z0=−1
...
1423
Accept the null hypothesis
No,at 0
...

Explanation:
Ho:μ=143Ha:μ<143
values are given as ,
α=0
...
07
p-value=0
...
1423 > 0
...
05 significance level, the data does not provide sufficient evidence to conclude
that Marie's mean game score is less than 143
...
Your hard work is paying off 😀😀
...
1423 is greater than the significance
level α=0
...

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
...
If α≤p-value, do not reject H0
...
In this case, α=0
...
1423, so the decision is to not reject the null hypothesis

---