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Part 1
1. Explain the difference between a 95% confidence interval and a 99% confidence interval
in terms of probability.
A 95% confidence interval would be narrower than a 99% confidence interval
a) To construct a 95% confidence interval for a population mean µ, what is the correct
critical value z*?
=

=

= 1.96
b) To construct a 99% confidence interval for a population mean µ, what is the correct
critical value z*?
=

=

= 2.58
2. Explain what the margin of error is and how to calculate it.
Margin of error(MOE) is the range of values lying above and below population mean in
confidence interval.
MOE =
󰇡
󰇢
Where:
is the z-score
is the sample size
is the population standard deviation
3. A survey of a group of students at a certain college, we call College ABC, asked: “About
how many hours do you study in a week?” The mean response of the 400 students is 15.8
hours. Suppose that the study time distribution of the population is known to be normal
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with a standard deviation of 8.5 hours. Use the survey results to construct a 95%
confidence interval for the mean study time at the College ABC.
N=400
=8.5
= 15.8
 = (1-0.95) = 0.05
=
=

=

= 1.96
MOE =

󰇡
󰇢 = 1.96󰇡


󰇢 = 0.833
95% Confidence interval limits are given as:
Lower limit =
MOE = 15.8 0.833 = 14.967
Upper limit =
+MOE = 15.8 + 0.833 = 16.633
Therefore,
95% CI = (14.967, 16.633)
14.96716.633
4. Explain the difference between a null hypothesis and an alternative hypothesis.
Null hypothesis is a statement about a population parameter on test in attempt to find
numerical evidence against it. In most cases, null hypothesis has a equality sign ().
Alternative hypothesis states a claim against null hypothesis statement. In nutshell,
alternative hypothesis is an assertion that holds if the null hypothesis is false. In most
cases, alternative hypothesis contains an inequality sign () and it determines
the direction of the hypothesis.
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5. Suppose that you are testing a null hypothesis H
0
: µ = 10 against the alternate H
1
: µ ≠ 10.
A simple random sample of 35 observations from a normal population is used for a test.
What values of the z statistic are statistically significant at the = 0.05 level?
N=35
The hypothesis is two-sided because the alternative hypothesis states that the population
is not equal to 10.
Therefore the z score at 0.05 significance level is calculated as:
=
=

=

= 1.96
Hence all the values of z statistic which does not belong to -1.96 and 1.96 the results are
statistically significant
Answer: All values of Z statistic except (-1.96, 1.96)
6. Describe the four-step process for tests of significance according to the textbook.
i. State null hypothesis and alternative hypothesis. The alternative
hypothesis should seek to answer the question at hand while null
hypothesis should contradict the alternative hypothesis.
ii. Choose the significance level. The alpha value chosen should be at a
low level that is 0.05, 0.01, and 0.1.
iii. Choose the right test statistic and determine the critical region based
on the significance level.
iv. Reject or fail to reject null hypothesis based on the test statistic result.
If the test result lies outside the critical region, the reject null
hypothesis otherwise fail to reject null hypothesis. Finally, make
conclusions based on the statistical evidence.
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Part 2
1. A study of a group of 40 male league bowlers chosen at random had an average score
was 176. It is known that the standard deviation of the population is 9.
a) Construct the 95% confidence interval for the mean score of all league bowlers.
N=40
=9
= 176
 = (1-0.95) = 0.05
=
=

=

= 1.96
MOE =

󰇡
󰇢 = 1.96󰇡

󰇢 = 2.789
95% Confidence interval limits are given as:
Lower limit =
MOE = 176 2.789= 173.211
Upper limit =
+MOE = 176 + 2.789= 178.789
Therefore,
95% CI = (173.211, 178.789)
b) Construct the 95% confidence interval for the mean score of all league bowlers
assuming that a sample of size 100 is used instead of 40, and the same mean and standard
deviation occur.
N=100
=9
= 176
 = (1-0.95) = 0.05
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=
=

=

= 1.96
MOE =

󰇡
󰇢 = 1.96󰇡

󰇢 = 1.764
95% Confidence interval limits are given as:
Lower limit =
MOE = 176 1.764= 174.236
Upper limit =
+MOE = 176 + 1.764= 177.764
Therefore,
95% CI = (174.236, 177.764)
c) Give the margin of error for each interval.
When n=40;
Margin of error = 2.789
When n=100;
Margin of error = 1.764
d) Explain why one confidence interval is larger than the other.
The sample size decreased margin of error because the margin of error is inversely
proportional to sample size.
2. There are 100 apartments in certain a San Francisco apartment building. The owner of the
building wants to estimate the mean number of people living in an apartment. The owner
draws a random sample of 40 apartments in the building. The number of people living in
each apartment is as follows:
1 2 1 2 3 1 3 4 3 1
2 2 1 2 2 2 1 3 2 3
2 3 1 2 3 3 2 4 5 2
3 2 2 3 1 1 2 2 1 2
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a. Compute the sample mean and sample standard deviation.
Sample mean,
=
= 2.175
Sample standard deviation, s =
󰇛

󰇜

= 0.958
b) Use the results from part (a) to construct a 95% confidence interval.
 = (1-0.95) = 0.05
Since the population standard deviation is unknown, we use t-statistic
Degrees of freedom, df = n-1 = 40-1 = 39
=
=

=

= 2.023
MOE =

󰇡
󰇢 = 󰇡


󰇢 = 0.306
95% Confidence interval limits are given as:
Lower limit =
MOE = 2.175 0.306= 1.869
Upper limit =
+MOE = 2.175 + 0.306 = 2.481
Therefore,
95% CI = (1.869, 2.481)
3. A doctor wishes to estimate the birth weights of infants. How large a sample must the
doctor select if she desires to be 99% confident that the true population mean is at most 6
ounces away from the mean of the sample? Assume the standard deviation is 8 ounces.
Hint: The margin of error should be at most 6.
= 1% = 0.01
=


=
=

= 2.58
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= 8
Margin of error, MOE = 6
n=󰇩


󰇪
= 󰇣

󰇤
=11.8 12
n=12
4. In Problem 4 of Part 1 a class survey of 400 students was given in which students at
College ABC claimed to study an average of 15.8 hours per week. Consider these
students as a simple random sample from the particular population of College ABC
students. We want to investigate the question: Does the survey provide good evidence
that students study more than 15 hours per week on average? Assume the population of
hours studied is normal with a standard deviation of 4.
Before working out this problem, it will help to look over the webpage, Hypothesis tests for
means, http://stattrek.com/hypothesis-test/mean.aspx?Tutorial=AP
a) State the null and alternate hypothesis in terms of the mean study time in hours for the
population.
Null hypothesis,
: = 15
Alternative hypothesis,
: > 15
b) Is this a one-tailed test or two-tailed test?
One-tailed test
c) Determine the value of the test statistic.
Test statistic = z
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Z=

󰇡
󰇢
=


= 4
Z=4
d) Sketch a normal shape curve and identify the test statistic.
e) Indicate the p-value of the test. Use the standard normal table. Shade the area under the
normal curve corresponding to the p-value. You can also use the website cited above to do this.
Z=4 (right tailed)
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The associated p-value = 0
P-value = 4 (from standard normal table)
f) State your conclusion to the statistical problem in terms of the null hypothesis, and your
conclusion to the practical problem.
P-value = 0
= 0.05
Since p-value (0) < 0.05, the results are statistically at 5% level of significant therefore we reject
null hypothesis.
Conclusion: At = 0.05, there is evidence that students study more than 15 hours per week on
average.

Unformatted Attachment Preview

Part 1 1. Explain the difference between a 95% confidence interval and a 99% confidence interval in terms of probability. A 95% confidence interval would be narrower than a 99% confidence interval a) To construct a 95% confidence interval for a population mean µ, what is the correct critical value z*? 𝑍𝛼 = 𝑍0.05 = 𝑍0.025 = 1.96 2 2 b) To construct a 99% confidence interval for a population mean µ, what is the correct critical value z*? 𝑍𝛼 = 𝑍0.01 = 𝑍0.005 = 2.58 2 2 2. Explain what the margin of error is and how to calculate it. Margin of error(MOE) is the range of values lying above and below population mean in confidence interval. 𝜎 MOE = 𝑍𝛼 ( 𝑛) 2 √ Where: 𝑍𝛼 is the z-score 2 𝑛 is the sample size 𝜎 is the population standard deviation 3. A survey of a group of students at a certain college, we call College ABC, asked: “About how many hours do you study in a week?” The mean response of the 400 students is 15.8 hours. Suppose that the study time distribution of the population is known to be normal with a standard deviation of 8.5 hours. Use the survey results to construct a 95% confidence interval for the mean study time at the College ABC. N=400 𝜎=8.5 𝑋̅ = 15.8 significance level = (1-0.95) = 0.05 critical value = Zα = 𝑍0.05 = 𝑍0.025 = 1.96 2 2 𝜎 MOE = 𝑍0.025 ( 𝑛) = 1.96( √ 8.5 ) = 0.833 √400 95% Confidence interval limits are given as: Lower limit = 𝑋̅ –MOE = 15.8 – 0.833 = 14.967 Upper limit = 𝑋̅ +MOE = 15.8 + 0.833 = 16.633 Therefore, 95% CI = (14.967, 16.633) 14.967< 𝜇 <16.633 4. Explain the difference between a null hypothesis and an alternative hypothesis. Null hypothesis is a statement about a population parameter on test in attempt to find numerical evidence against it. In most cases, null hypothesis has a equality sign (≥, ≤, =). Alternative hypothesis states a claim against null hypothesis statement. In nutshell, alternative hypothesis is an assertion that holds if the null hypothesis is false. In most cases, alternative hypothesis contains an inequality sign (<, > 𝑜𝑟 ≠) and it determines the direction of the hypothesis. 5. Suppose that you are testing a null hypothesis H0: µ = 10 against the alternate H1: µ ≠ 10. A simple random sample of 35 observations from a normal population is used for a test. What values of the z statistic are statistically significant at the = 0.05 level? N=35 The hypothesis is two-sided because the alternative hypothesis states that the population is not equal to 10. Therefore the z score at 0.05 significance level is calculated as: critical value = Zα = 𝑍0.05 = 𝑍0.025 = ±1.96 2 2 Hence all the values of z statistic which does not belong to -1.96 and 1.96 the results are statistically significant Answer: All values of Z statistic except (-1.96, 1.96) 6. Describe the four-step process for tests of significance according to the textbook. i. State null hypothesis and alternative hypothesis. The alternative hypothesis should seek to answer the question at hand while null hypothesis should contradict the alternative hypothesis. ii. Choose the significance level. The alpha value chosen should be at a low level that is 0.05, 0.01, and 0.1. iii. Choose the right test statistic and determine the critical region based on the significance level. iv. Reject or fail to reject null hypothesis based on the test statistic result. If the test result lies outside the critical region, the reject null hypothesis otherwise fail to reject null hypothesis. Finally, make conclusions based on the statistical evidence. Part 2 1. A study of a group of 40 male league bowlers chosen at random had an average score was 176. It is known that the standard deviation of the population is 9. a) Construct the 95% confidence interval for the mean score of all league bowlers. N=40 𝜎=9 𝑋̅ = 176 significance level = (1-0.95) = 0.05 critical value = Zα = 𝑍0.05 = 𝑍0.025 = 1.96 2 𝜎 2 MOE = 𝑍0.025 ( 𝑛) = 1.96( √ 9 ) = 2.789 √40 95% Confidence interval limits are given as: Lower limit = 𝑋̅ –MOE = 176 – 2.789= 173.211 Upper limit = 𝑋̅ +MOE = 176 + 2.789= 178.789 Therefore, 95% CI = (173.211, 178.789) b) Construct the 95% confidence interval for the mean score of all league bowlers assuming that a sample of size 100 is used instead of 40, and the same mean and standard deviation occur. N=100 𝜎=9 𝑋̅ = 176 significance level = (1-0.95) = 0.05 critical value = Zα = 𝑍0.05 = 𝑍0.025 = 1.96 2 2 𝜎 9 MOE = 𝑍0.025 ( 𝑛) = 1.96( ) = 1.764 √100 √ 95% Confidence interval limits are given as: Lower limit = 𝑋̅ –MOE = 176 – 1.764= 174.236 Upper limit = 𝑋̅ +MOE = 176 + 1.764= 177.764 Therefore, 95% CI = (174.236, 177.764) c) Give the margin of error for each interval. When n=40; Margin of error = 2.789 When n=100; Margin of error = 1.764 d) Explain why one confidence interval is larger than the other. The sample size decreased margin of error because the margin of error is inversely proportional to sample size. 2. There are 100 apartments in certain a San Francisco apartment building. The owner of the building wants to estimate the mean number of people living in an apartment. The owner draws a random sample of 40 apartments in the building. The number of people living in each apartment is as follows: 1 2 1 2 3 1 3 4 3 1 2 2 1 2 2 2 1 3 2 3 2 3 1 2 3 3 2 4 5 2 2 2 3 1 1 2 2 3 1 2 a. Compute the sample mean and sample standard deviation. Sample mean, 𝑋̅ = ∑ 𝑋𝑖 𝑛 = 2.175 Sample standard deviation, s = ∑(𝑋𝑖 −𝑋̅) 𝑛−1 = 0.958 b) Use the results from part (a) to construct a 95% confidence interval. significance level = (1-0.95) = 0.05 Since the population standard deviation is unknown, we use t-statistic Degrees of freedom, df = n-1 = 40-1 = 39 critical value = t α = 𝑡0.05 = 𝑡0.025 = 2.023 2 2 𝑠 0.958 MOE = 𝑡0.025 ( 𝑛) = 2.023 ( √ √40 ) = 0.306 95% Confidence interval limits are given as: Lower limit = 𝑋̅ –MOE = 2.175 – 0.306= 1.869 Upper limit = 𝑋̅ +MOE = 2.175 + 0.306 = 2.481 Therefore, 95% CI = (1.869, 2.481) 3. A doctor wishes to estimate the birth weights of infants. How large a sample must the doctor select if she desires to be 99% confident that the true population mean is at most 6 ounces away from the mean of the sample? Assume the standard deviation is 8 ounces. Hint: The margin of error should be at most 6. 𝛼= 1% = 0.01 𝛼 2 = 0.01 2 = 0.005 Critical value = Zα = 𝑍0.005 = 2.58 2 𝜎= 8 Margin of error, MOE = 6 Zα ∗𝜎 2 2 n=[ 𝑀𝑂𝐸 ] =[ 2.58∗8 2 6 ] =11.8 ≈12 n=12 4. In Problem 4 of Part 1 a class survey of 400 students was given in which students at College ABC claimed to study an average of 15.8 hours per week. Consider these students as a simple random sample from the particular population of College ABC students. We want to investigate the question: Does the survey provide good evidence that students study more than 15 hours per week on average? Assume the population of hours studied is normal with a standard deviation of 4. Before working out this problem, it will help to look over the webpage, Hypothesis tests for means, http://stattrek.com/hypothesis-test/mean.aspx?Tutorial=AP a) State the null and alternate hypothesis in terms of the mean study time in hours for the population. Null hypothesis, 𝐻0 : 𝜇 = 15 Alternative hypothesis, 𝐻1 : 𝜇 > 15 b) Is this a one-tailed test or two-tailed test? One-tailed test c) Determine the value of the test statistic. Test statistic = z Z= 𝑋̅ −𝜇 ( 𝜎 ) √𝑛 = 15.8−15 4 √400 =4 Z=4 d) Sketch a normal shape curve and identify the test statistic. e) Indicate the p-value of the test. Use the standard normal table. Shade the area under the normal curve corresponding to the p-value. You can also use the website cited above to do this. Z=4 (right tailed) The associated p-value = 0 P-value = 4 (from standard normal table) f) State your conclusion to the statistical problem in terms of the null hypothesis, and your conclusion to the practical problem. P-value = 0 𝛼= 0.05 Since p-value (0) < 0.05, the results are statistically at 5% level of significant therefore we reject null hypothesis. Conclusion: At 𝛼= 0.05, there is evidence that students study more than 15 hours per week on average. Name: Description: ...
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