Download 1 - JustAnswer

1. A sample of 50 golfers showed that their average score on a particular golf course was
90 with a standard deviation of 10. Answer each of the following (show all work): (A)
Find the 99% confidence interval of the mean score for all 50 golfers. (B) Find the 99%
confidence interval of the mean score for all golfers if this is a sample of 80 golfers
instead of a sample of 50. (C) Which confidence interval is larger and why?
The 99% confidence interval for the mean score is given by,
x  t / 2 s / n , x  t / 2 s / n ,


where t / 2 is obtained from Student’s t distribution with (n-1) degrees of freedom.
A) Here it is given that, n = 50, x = 90 and s = 10. Therefore, t / 2 = 2.68
Thus, the 99% confidence interval for the mean score is given by
(90 – 2.68*10/√50, 90 – 2.68*10/√50)
= (90 – 3.79, 90+3.79)
= (86.21, 93.79)
B) Here it is given that, n = 80, x = 90 and s = 10. Therefore, t / 2 = 2.639
Thus, the 99% confidence interval for the mean score is given by
(90 – 2.639*10/√80, 90 – 2.639*10/√80)
= (90 – 2.95, 90+2.95)
= (87.05, 92.95)
C) The first confidence interval is larger because the standard error increases as the
sample size decreases.
2. Consider the sample data below. 34 21 23 52 34 38 36 49 36 34 20 50 44 36 43 27 45
40 52 39 31 54 20 30 34 50 40 25 26 27 26 21 43 Suppose we wish to test the following
hypotheses for the data: Ho:u 40 Given the mean and the standard deviation of the data as
above: (A) Compute the appropriate test statistic for testing Ho. (show all work). (B)
Determine the critical value for a = 0.05. (C) Test Ho using the traditional/classical
method. (D) State the decision based on the results of the test. (E) Had you used the pvalue method of hypothesis testing, what would the p-value have been?
A) The test statistic for testing Ho is
x  40
, follows a Student’s t distribution with (n-1) degrees of freedom.
t
s/ n
From the given data,
n = 33, x = 35.7576 and s = 10.2684,
Therefore, the test statistic is given by,
35.7576  40
= -2.3734
t
10.2684 / 33
B) If the alternative hypothesis is Ha:   40, then the given test is a two-tailed t-test.
Since a = 0.05, from Student’s t distribution with (n-1) = 32 degrees of freedom, the
critical value of the test is 2.037.
If the alternative hypothesis is Ha:   40, then the given test is a one-tailed t-test.
Since a = 0.05, from Student’s t distribution with (n-1) = 32 degrees of freedom, the
critical value of the test is -1.694.
C) For the two tailed t-test, the critical region of the test is |t| > 2.037.
Here, |t| = 2.3734 > 2.037.
Therefore we reject the null hypothesis Ho.
For the on-tailed t-test, the critical region of the test is t < -1.694.
Here, t = -2.3734 < -1.694.
Therefore we reject the null hypothesis Ho.
D) Decision: We reject the null hypothesis Ho:  =40
E) For the two-tailed t test, the p-value is given by
P[|t|>2.3734] = 0.0238 (Using t distribution with 32 d.f.)
Since the p-value < 0.05, we reject the null hypothesis Ho:  =40.
For the one-tailed t test, the p-value is given by
P[ t < -2.3734] = 0.0119 (Using t distribution with 32 d.f.)
Since the p-value < 0.05, we reject the null hypothesis Ho:  =40.
3. A researcher is interested in estimating the noise levels in decibels at area urban
hospitals. She wants to be 90% confident that her estimate is correct. If the standard
deviation is 4.8, how large a sample is needed to get the desired information to be
accurate within 0.70 decibels? Show all work.
Here it is given that,
 = 4.8, E = 0.70, z / 2 = 1.645
Now, the required sample size,
 z    1.645* 4.8 
n =   /2  = 
 =127.2157 = 128
 E   0.70 
Thus the required sample size = 128
2
2
4. A sample size of 30 is used to test Ho:u=35 vs Ha:u ?. Given that x mean 35.9 and ? =
4.5, answer the following questions: (A) What is the compute value of the test statistic?
(B) What distribution does the test statistic have when the null hypothesis is true? (C) Is
the alternative hypothesis one-tailed or two-tailed? (D) What is the p-value?
A) The test statistic for testing Ho is
x  35
t
s/ n
Here it is given that,
n = 30, x = 35.9 and s = 4.5,
Therefore, the test statistic is given by,
35.9  35
= 1.0954
t
4.5 / 30
B) When the null hypothesis is true, the test statistic follows a Student’s t distribution
with (n-1) = (30-1) = 29 degrees of freedom.
{I assume that the sample standard deviation s is given as 4.5. If it is the population
standard deviation the test statistic follows a Standard normal distribution.}
C) If the alternative hypothesis is Ha: u ≠ 35, then it is a two-tailed test.
If the alternative hypothesis is Ha: u > 35, then it is a one-tailed test.
D) If the alternative hypothesis is Ha: u ≠ 35, the p-value of the test is given by,
p-value =P[|t| > 1.0954] = 0.2823 (Using t distribution with 29 df)
If the alternative hypothesis is Ha: u > 35, the p-value of the test is given by,
p-value =P[ t > 1.0954] = 0.1412 (Using t distribution with 29 df)
{ If the population standard deviation σ is given in this problem, the test statistic follows
a Standard normal distribution and the p-values are computed as follows
If the alternative hypothesis is Ha: u ≠ 35, the p-value of the test is given by,
p-value =P[|Z| > 1.0954] = 0.2733 (Using Standard Normal distribution)
If the alternative hypothesis is Ha: u > 35, the p-value of the test is given by,
p-value =P[ Z > 1.0954] = 0.1367 (Using Standard Normal distribution) }
5. A researcher claims that the average age of people who buy lottery tickets is 60. A
sample of 30 is selected and their ages are recorded as shown below. The standard
deviation is 15. At a = 0.01 is there enough evidence to reject the researcher’s claim?
Show all work. 65 63 75 52 22 80 72 56 82 56 24 60 70 74 70 61 65 71 39 74 79 75 71
49 62 68 71 67 69 45
Let  denote the average age of people who buy lottery tickets.
Here we want to test the null hypothesis Ho:  = 60 against the alternative hypothesis
Ha:  ≠ 60.
The test statistic for testing Ho is
x  60
, follows a standard Normal distribution.
z
/ n
Here it is given that,
n = 30,  = 15 and x = 62.9
Thus the test statistic is given by
62.9  60
=1.0589
z
15 / 30
Since a = 0.01, the critical value of the test is 2.58.
Thus, the critical region of the test is |z| > 2.58
Here, |z| = 1.0589 < 2.58
Therefore, we fail to reject the null hypothesis Ho.
So at a = 0.01, there is not enough evidence to reject the researcher’s claim.
6. Given a level of confidence of 95% and a population standard deviation of 8, what
other information is necessary: (A) To find the Maximum Error of Estimate (E)? (B) To
find the sample size (n)? (C) Given the above confidence level and population standard
deviation, find the Maximum Error of Estimate (E) if n = 45. Show all your calculations.
Show all work. (D) For this same sample of n = 45, what is the width of the confidence
interval around the population mean? Show all work. (E) Given this same confidence
level and standard deviation, find n if E = 3.1. (Always round to the nearest whole
person.) Show all work.
A) To find the Maximum Error of Estimate (E), the sample size (n) is also required in
addition to the population standard deviation(σ) and Z value associated with the given
degree of confidence( z / 2 ).
B) To find the sample size (n), the Maximum Error of Estimate (E) is also required in
addition to the population standard deviation(σ) and Z value associated with the given
degree of confidence( z / 2 ).
z 
C) The Maximum Error of Estimate, E =  / 2
n
Here it is given that,
 = 8, n = 45, z / 2 = 1.96
Thus,
1.96*8
E=
= 2.3374
45
D) The width of the confidence interval around the population mean is
2*E = 2*2.3374 = 4.6748
Note that the confidence interval around the population mean ( x - E, x + E)
E)
The required sample size,
z 
n =   /2 
 E 
Here it is given that,
 = 8, E = 3.1, z / 2 = 1.96
So, the required sample size,
2
2
 1.96*8 
n= 
 = 25.583 = 26
 3.1 
Thus the required sample size is 26.