Strayer University - MAT 300 Unit 5 Milestone 5 Exam. Questions And Answers.Mrs. Pellegrin has weighed 5 packages of cheese and recorded the weights as
... [Show More] 10.2 oz, 10.5 oz, 9.3 oz, 9.8 oz, and 10.0 oz. She calculated the
standard deviation to be 0.45 oz.
Select the 95% confidence interval for Mrs. Pellegrin's set of data.
9.4 to 10.52
9.53 to 10.39
9.34 to 10.44
9.48 to 10.44
RATIONALE
In order to get the 95% CI , we first need to find the critical t-score. Using a t-table, we need to find (n-1) degrees of freedom, or (5-1) = 4 df and
the corresponding CI
Using the 95% CI in the bottom row and 4 df on the far left column, we get a t-critical score of 2.776.
We also need to calculate the mean:
So we use the formula to find the confidence interval:
The lower bound is:
9.96 -0.56 = 9.40
The upper bound is:
9.96+0.56 = 10.52
CONCEPT
Confidence Intervals Using the T-Distribution
2
A school is gathering some data on its sports teams because it was believed that the distribution of boys and girls were evenly distributed across all
the sports. This table lists the number of boys and girls participating in each sport.
Boys Girls
Tennis 18 30
Soccer 42 15
Swimming 12 18
Select the observed and expected frequencies for the boys participating in soccer.
Observed: 42
Expected: 24
Observed: 42
Expected: 22.5
Observed: 57
Expected: 22.5
Observed: 57
Expected: 24
RATIONALE
If we simply go to the chart then we can directly see the observed frequency for boys participating in soccer is 42.
To find the expected frequency, we need to find the number of
occurrences if the null hypothesis is true, which in this case, was that the three options are equally likely, or if the three options were all evenly
distributed.
First, add up all the options in the boys column:
If each of these three options were evenly distributed among the 72 boys, we would need to divide the total evenly between the three options:
This means we would expect 24 boys to choose tennis, 24 boys to choose soccer, and 24 boys to choose swimming.
CONCEPT
Chi-Square Statistic
3
What value of z* should be used to construct a 98% confidence interval of a population mean? Answer choices are rounded to the
thousandths place.
1.175
2.055
1.645
2.325
RATIONALE
Using the z-chart to construct a 98% CI, this means that there is 1% for each tail. The lower tail would be at 0.06 and the upper tail would be at (1
- 0.01) or 0.99. The closest to 0.94 on the z-table is between 0.9901 and 0.9898.
0.9898 corresponds with a z-score of 2.32.
0.9901 corresponds with a z-score of 2.33.
Taking the average of these two scores, we get a z-score of 2.325.
CONCEPT
Confidence Intervals
4
A table represents the number of students who passed or failed an aptitude test at two different campuses.
East Campus West Campus
Passed 39 45
Failed 61 55
In order to determine if there is a significant difference between campuses and pass rate, the chi-square test for association and independence
should be performed.
What is the expected frequency of East Campus and passed?
48.3 students
50.5 students
42 students
39 students
RATIONALE
In order to get the expected counts we can note the formula is:
CONCEPT
Chi-Square Test for Homogeneity
5
Joe hypothesizes that the average age of the population of Florida is less than 37 years. He records a sample mean equal to 37 and states the
hypothesis as μ = 37 vs μ < 37.
What type of test should Joe do?
Left-tailed test
Right-tailed test
Joe should not do any of the types of tests listed
Two-tailed test
RATIONALE
Since the Ha is a less than sign, this indicates he wants to run a 1
tailed test where the rejection region is the lower or left tail.
This can be called a left-tailed test.
CONCEPT
One-Tailed and Two-Tailed Tests
6
Select the statement that correctly describes a Type II error.
A Type II error occurs when the null hypothesis is rejected when it is actually true.
A Type II error occurs when the null hypothesis is rejected when it is actually false.
A Type II error occurs when the null hypothesis is accepted when it is actually false.
A Type II error occurs when the null hypothesis is accepted when it is actually true.
RATIONALE
Recall a Type II error is when we incorrectly accept a false null
hypothesis. In this case, we want to reject and conclude there is evidence is correct.
CONCEPT
Type I/II Errors
7
The data below shows the heights in inches of 10 students in a class.
Student Height, in inches
Student 1 53
Student 2 52.5
Student 3 54
Student 4 51
Student 5 50.5
Student 6 49.5
Student 7 48
Student 8 53
Student 9 52
Student 10 50
The standard error of the sample mean for this set of data is __________. Answer choices are rounded to the hundredths place.
0.19
1.77
1.87
0.59
RATIONALE
In order to get the standard error of the mean, we can use the following formula:
fraction numerator s over denominator square root of n end fraction, where is the standard deviation and is the sample size.
Either calculate by hand or use Excel to find the standard deviation, which is 1.87. The sample size is 10 students.
The standard error is then:
CONCEPT
Calculating Standard Error of a Sample Mean
8
One condition for performing a hypothesis test is that the observations are independent. Mark is going to take a sample from a population of 400
students.
How many students will Mark have to sample without replacement to treat the observations as independent?
360
40
80
300
RATIONALE
In general we want about 10% or less to still assume independence.
So size = 0.1*N = 0.1(400) = 40
A sample of 40 or less would be sufficient.
CONCEPT
Sampling With or Without Replacement
9
Mrs. Pellegrin has weighed 5 packages of cheese and recorded the weights as 10.2 oz, 10.5 oz, 9.3 oz, 9.8 oz, and 10.0 oz. She calculated the
standard deviation to be 0.45 oz.
The cheese factory lists the package weight at 10 oz.
The t-statistic for a two-sided test would be __________.
+0.20
-0.20
+0.40
-0.40
RATIONALE
Using the information given, we need to find the sample mean:
We now know the following information:
Let's plug in the values into the formula:
CONCEPT
T-Tests
10
Rachel recorded the number of calls she made at work during the week:
Day Calls
Monday 18
Tuesday 14
Wednesday 24
Thursday 16
She expected to make 18 calls each day. To answer whether the number of calls follows a uniform distribution, a chi-square test for goodness of fit
should be performed. (alpha = 0.10)
What is the chi-squared test statistic? Answers are rounded to the nearest hundredth.
3.11
4.61
3.27
1.52
RATIONALE
Using the chi-square formula we can note the test statistic is [Show Less]