1 Math 107 – Modern Elementary Statistics Extra Credit/Remedial Work Instruction: Answer all questions! Show your work properly! PART 2 Name of Student:...

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1 Math 107 – Modern Elementary Statistics Extra Credit/Remedial Work Instruction: Answer all questions! Show your work properly! PART 2 Name of Student: ____________________________________________________ 1. (8 Points) Find the critical value Zα/2 corresponding to the given confidence level. Confidence Level α Critical Value, Zα/2 92% 95% 94% 98% Answers: Show your work here. 2. (5 Points) The Medical Rehabilitation Education Foundation(MREF) reports that the average cost of rehabilitation for stroke victims is $24,672. To see if the average cost of rehabilitation is different at a particular hospital, a researcher selects a random sample of 35 stroke victims at the hospital and finds that the average cost of their rehabilitation is $25,226. The standard deviation of the population is $3251. At α = 0.01, can it be concluded that the average cost of stroke rehabilitation at a particular hospital is different from $24,672? Answers: 3. (8 Points) A popular social media platform claims that on an average each of its registered users logs into their profiles for about 30 times in a year. If the number of login data for a random sample of 144 users is collected and analyzed, it is found that the average login count in a year is 29. Assume that the population standard deviation is 3.1 Calculate the 95% confidence interval for population mean using the sample mean. Is the company’s claim is believable? Answer(s): 4. (5 Points) Assume the time taken by racers to finish the running race (100 mts) follows a normal distribution. If you want to estimate the average time taken by racers to finish the 100 mts race with 90% confidence level and you want to express the interval within 2 secs, what is the minimum sample size that you need to consider? From the past studies the standard deviation for time taken by racers is 2.1 secs. Answer(s): 5. (5 Points) A research firm wants to estimate the average time (Industry average) of a client call. A random sample of 60 calls made by sales teams from different companies is analyzed. Average time for the sample is found to be 3.4 minutes. If the standard deviation of population is known to be 0.4 minutes, Estimate the 99% confidence interval for the average time of client call and interpret this interval. Answer(s): 6. (6 Points) If 10% of the bolts produced by a machine are defective, determine the probability that out of 5 bolts chosen at random: a. 2 will be defective b. None(that is, 0) will be defective c. At most 2 bolts will be defective. Answer(s): 7. (5 Points) It is claimed that a particular flu shot is effective in 80% of the cases. If a group of 15 adults receive the flu shot, find the probability that exactly 11 will benefit from the shot. Answer(s): 8. (10 Points) The number of televisions per household in a small town is shown on the table below: # of Televisions(X) # of Households 0 26 1 442 2 728 3 1404 a) Use the information in above table to construct a probability distribution. b) Graph the probability distribution using a histogram and describe its shape. c) Find the mean, variance, and standard deviation of the probability distribution. d) Interpret the results. Answer(s): 9. (10 Points) A 911 service center recorded the number of calls received per hour. The random variable X represents the number of calls per hour for one week. # of Calls (X) Probability, P(X) 0 0.01 1 0.10 2 0.26 3 0.33 4 0.18 5 0.06 6 0.03 7 0.03 a) Find the mean, variance and standard deviation of this probability distribution. b) Graph this probability distribution using a histogram and describe its shape b) Find the expected value of this probability distribution c) Interpret the results. Answer(s): 10. (6 Points) Find the mean, variance and standard deviation for each of the values of n and p. a. n = 166, p = 0.75 b. n = 28, p = 0.86 c. n = 50, p = 3/5 Answer(s): 11. (15 Points) Find the probabilities for each, using the standard normal distribution. a. P(0 ˂ Z ˂ 1.31) f. P(1.92 ˂ Z ˂ 2.44) b. P(Z ˂ 2.11) g. P(-0.63 ˂ Z ˂-1.52) c. P(Z > 2.45) d. P(Z ˂ -1.32) e. P(- 2.33 ˂ Z ˂1.76) Answer(s): 12. (4 Points) If the standard deviation of a national accounting examination is 30, how large a sample is needed to estimate the true mean score within 5 points with 99% confidence? Answers: 13. (10 Points) Find the critical value Zα/2 and use your answers to complete the following table. Show all work. Confidence Level α Critical Value, Zα/2 92% 94% 97% 85% Answer(s): 14. (10 Points) A car company says that the mean mileage for its luxury sedan is at least 23 miles per gallon(mpg). You believe the claim is incorrect and find that a random sample of 5 cars has a mean gas mileage of 22 miles per gallon and a standard deviation of 4 mpg. At α = 0.05, test the company’s claim. Assume the population is normally distributed. Answer(s): 15. (4 Points) Which of the following is not a statistical hypothesis? (a) 100 m = (b) 100 m > (c) 0.5 x > (d) 100 s > Answer(s): 16. (4 Points) Which of the following pair of statistical hypotheses is valid? (a) 0 :150 vs. :150 a HH mm => (b) 0 :150 vs. :150 a HH mm <> (c) 0 :120 vs. :100 a HH mm == (d) 0 :120 vs. :100 a HH mm =¹ Answer(s): 17. (4 Points) Which of the following statement is true? (a) The null hypothesis, denoted by o H , is the claim that is initially assumed to be true in the hypothesis testing. (b) The alternative hypothesis, denoted by a H , is the assertion that is contradictory to the null hypothesis o H . (c) The null hypothesis o H will be rejected in favor of the alternative hypothesis only if sample evidence suggests that o H is false. (d) If sample evidence does not strongly contradict the null hypothesis o H , we will continue to believe in the truth of o H (e) All of the above statements are true Answer(s): 18. (4 Points) Let a = P(making a Type I error) and b = P(making a Type II error). The null hypothesis is rejected when (a) the p-value is larger than b (b) the p-value is smaller than b (c) the p-value is larger than a (d) the p-value is smaller than a Answer(s): 19. (4 Points) In hypothesis-testing analysis, a type II error occurs if the null hypothesis o H is a) Rejected when it is true b) Rejected when it is false. c) Not rejected when it is false d) Not rejected when it is true. Answer(s): GOOD LUCK! _1226744705.unknown _1226750773.unknown _1226750806.unknown _1226750790.unknown _1226750735.unknown _1226744666.unknown _1226744681.unknown _1226744694.unknown _1110024757.unknown _1226060604.unknown _1226060611.unknown _1226060538.unknown _1226060550.unknown _1133974642.unknown _1110024701.unknown
Answered 2 days AfterAug 07, 2021

Answer To: 1 Math 107 – Modern Elementary Statistics Extra Credit/Remedial Work Instruction: Answer all...

Suraj answered on Aug 08 2021
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1
Math 107 – Modern Elementary Statistics
Extra Credit/Remedial Work
Instruction: Answer all questions! Show your work properly!
PART 2
Name of Student: ____________________________________________________
1. (8 Points) Find the critical value Zα/2 corresponding to the given confidence level.
    Confidence Level
    α
    Critical Value, Zα/2
    92%
    
    
    95%
    
    
    94%
    
    
    98%
    
    
Answers:
    Confidence Level
    α
    Critical Value, Zα/2
    92%
    0.08
    1.75
    95%
    0.05
    1.96
    94%
    0.06
    1.88
    98%
    0.02
    2.32
6
Show your work here.
We can get the alpha,
a
value by subtracting the confidence level from 1. The calculations are given as follows:
92%0.92
10.92
0.08
Confidencelevel
a
==
=-
=
All the alpha values are calculated in same way.
To calculate the critical value, we need the normal distribution table or we can use MS-Excel formula.
For
0.08
a
=
,
0.08/2
NORM.S.INV(0.08/2)
z
=

1.75
=
In the same way we can calculate critical values by just changing the alpha value in the above formula.
2. (5 Points) The Medical Rehabilitation Education Foundation (MREF) reports that the average cost of rehabilitation for stroke victims is $24,672. To see if the average cost of rehabilitation is different at a particular hospital, a researcher selects a random sample of 35 stroke victims at the hospital and finds that the average cost of their rehabilitation is $25,226. The standard deviation of the population is $3251. At α = 0.01, can it be concluded that the average cost of stroke rehabilitation at a particular hospital is different from $24,672?
Answers:
Consider the given information,
The average cost of rehabilitation for stroke victims,
24672
m
=
The sample size, n = 35
The sample mean,
25226
x
=
The population standard deviation,
3251
s
=
The level of significance,
0.01
a
=
The hypotheses are given as follows:
0
1
:24672
:24672
H
H
m
m
=
¹
The test-statistic is given as follows:
/
x
z
n
m
s
-
=
(Because population standard deviation is given)
2522624672
3251/35
1.008
z
-
=
=
Since, the test is a two-tailed test. Thus, the p-value is calculated as follows:
(
)
21.008
20.1567
0.3134
pvaluePz
-=´>

=
Since, the p-value is greater than the level of significance (0.3134 > 0.05). Thus, the null hypothesis is not rejected. Hence, we do no have enough evidence against the alternative hypothesis.
3. (8 Points) A popular social media platform claims that on an average each of its registered users logs into their profiles for about 30 times in a year. If the number of login data for a random sample of 144 users is collected and analyzed, it is found that the average login count in a year is 29. Assume that the population standard deviation is 3.1
Calculate the 95% confidence interval for population mean using the sample mean. Is the company’s claim is believable?
Answer(s):
The sample size, n = 144
The sample mean,
29
x
=
The population standard deviation,
3.1
s
=
The significance level,
0.05
a
=
The
/2
z
a
critical value from normal distribution table is 1.96.
The formula for the confidence interval is given as follows:
/2/2
,
xzxz
nn
aa
ss
æö
-+
ç÷
èø
Substitute the values in the above formula,
(
)
3.13.1
291.96,291.96
144144
29.49,29.51
æö
-´+´
ç÷
èø
Thus, 95% confidence interval is 29.49 and 29.51. The confidence interval does not contain the true population mean 30. Thus, the company claim is claim is correct.
4. (5 Points) Assume the time taken by racers to finish the running race (100 mts) follows a normal distribution. If you want to estimate the average time taken by racers to finish the 100 mts race with 90% confidence level and you want to express the interval within 2 secs, what is the minimum sample size that you need to consider? From the past studies the standard deviation for time taken by racers is 2.1 secs.
Answer(s):
The objective is to calculate the sample size.
The information given is as follows:
Confidence level = 90%
Significance level,
10.90
a
=-

0.10
=
0.10/2
1.645
z
=
The population standard deviation, s = 2.1
Margin of error = 2
The required sample size,
2
/2
z
n
a
s
e
´
æö
=
ç÷
èø

2
1.6452.1
2
3
´
æö
=
ç÷
èø
;
5. (5 Points) A research firm wants to estimate the average time (Industry average) of a client call. A random sample of 60 calls made by sales teams from different companies is analyzed. Average time for the sample is found to be 3.4 minutes. If the standard deviation of population is known to be 0.4 minutes, Estimate the 99% confidence interval for the average time of client call and interpret this interval.
Answer(s):
The sample size, n = 60
The sample mean,
3.4
x
=
The population standard deviation,
0.4
s
=
The significance level,
0.01
a
=
The
/2
z
a
critical value from normal distribution table is 2.576.
The formula for the confidence interval is given as follows:
/2/2
,
xzxz
nn
aa
ss
æö
-+
ç÷
èø
Substitute the values in the above formula,
(
)
0.40.4
3.42.576,3.42.576
6060
3.27,3.53
æö
-´+´
ç÷
èø
Thus,...
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