MATH 300 / MAT300 STATISTICS
UNIT 5 MILESTONE 5
19 questions were answered correctly. 5 questions were answered incorrectly. 1
Adam tabulated the
... [Show More] values for the average speeds on each day of his road trip as 60.5, 63.2, 54.7, 51.6, 72.3, 70.7, 67.2, and 65.4 mph. The sample standard deviation is 7.309.
Select the 98% confidence interval for Adam’s set of data.
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46.94 to 79.46
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55.45 to 70.95
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46.94 to 71.33
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55.45 to 79.46
RATIONALE
In order to get the 98% CI , we first need to find the critical
t-score. Using a t-table, we need to find (n-1) degrees of freedom, or (8-1) = 7 df and the corresponding CI.
Using the 98% CI in the bottom row and 7 df on the far left column, we get a t-critical score of 2.998. We also need to calculate the mean:
So we use the formula to find the confidence interval:
The lower bound is:
63.2-7.75 = 55.45
The upper bound is: 63.2+7.75 = 70.95 CONCEPT
Confidence Intervals Using the T-Distribution
2
A table represents the number of students who passed or failed an aptitude test at two different campuses.
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 South Campus and passed?
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50 students
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36.5 students
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42 students
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43.7 students
RATIONALE
In order to get the expected counts we can note the formula is:
CONCEPT
Chi-Square Test for Homogeneity
3
Sukie interviewed 125 employees at her company and discovered that 21 of them planned to take an extended vacation next year.
What is the 95% confidence interval for this population proportion? Answer choices are rounded to the hundredths place.
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0.10 to 0.23
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0.16 to 0.17
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0.11 to 0.16
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0.11 to 0.21
RATIONALE
In order to get the CI we want to use the following form.
First, we must determine the corresponding z*score for 95% Confidence Interval. Remember, this means that we have 5% for the tails, meaning 5%, or 0.025, for each tail. Using a z-table, we can find the upper z-score by finding (1 - 0.025) or 0.975 in the table.
This corresponding z-score is at 1.96. We can know
So putting it all together:
The lower bound is:
0.168-0.065 =0.103 or 0.10
The upper bound is:
0.168+0.065 =0.233 or 0.23
CONCEPT
Confidence Interval for Population Proportion
4
A market research company conducted a survey of two groups of students from different schools. They found that students from school A spent an average of 90 minutes studying daily, while the students from school B spent an average of 75 minutes daily.
They want to find out if the difference in the mean times spent studying by the students of the two schools is statistically significant.
Which of the following sets shows the correct null hypothesis and alternative hypothesis?
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Null Hypothesis: There is no difference in the mean times spent by the schools' students. Alternative Hypothesis: There is at least some difference in the mean times spent by the schools' students.
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Null Hypothesis: There is at least some difference in the mean times spent by the schools' students. Alternative Hypothesis: The students from school B spend more time studying than the students from school A.
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Null Hypothesis: The difference in the mean times spent by the schools' students is 15 minutes. Alternative Hypothesis: There is no difference in the mean times spent by the schools' students.
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Null Hypothesis: School B students spend more time studying than School A.
Alternative Hypothesis: The difference in the mean times spent by the schools' students is 15 minutes.
RATIONALE
Recall that the null hypothesis is always of no difference.
So the null hypothesis (Ho) is that the mean time studying for group A = mean for group B. This would indicate no difference between the two groups.
The alternative hypothesis (Ha) is that there is difference in the mean study time between the two groups.
CONCEPT
Hypothesis Testing
5
Carl recorded the number of customers who visited his new store during the week:
He expected to have 15 customers each day. To answer whether the number of customers 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.
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0.40
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1.60
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0.67
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2.33
RATIONALE
Using the chi-square formula we can note the test statistic is
CONCEPT
Chi-Square Test for Goodness-of-Fit
6
Rachel measured the lengths of a random sample of 100 screws. The mean length was 2.9 inches, and the population standard deviation is 0.1 inch.
To see if the batch of screws has a significantly different mean length from 3 inches, what would the value of the z-test statistic be?
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1
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10
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-10
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-1
RATIONALE
If we first note the denominator of
Then, getting the z-score we can note it is
This tells us that 2.9 is 10 standard deviations below the value of 3, which is extremely far away.
CONCEPT
Z-Test for Population Means
7
What do the symbols , , and represent?
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Sample statistics
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Variables of interest
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Population parameters
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Defined variables
RATIONALE
Recall that is the sample proportion, is the sample mean, and is the sample standard
deviation. Since all of these come from samples they are statistics.
CONCEPT
Sample Statistics and Population Parameters
8
The data below shows the grams of fat in a series of popular snacks.
If Morris wanted to construct a one-sample t-statistic, what would the value for the degrees of freedom be?
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10
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9
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11
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5
RATIONALE
The degrees of freedom for a 1 sample t-test are df=n-1 where n is the sample size. In this case, n=10, then df = n-1 = 10-1 = 9.
CONCEPT
T-Tests
9
A researcher has a table of data with 5 column variables and 5 row variables.
The value for the degrees of freedom in order to calculate the statistic is .
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25
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24
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4
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16
RATIONALE
Recall to get the degrees of freedom we use df = (r-1)(c-1) where c and r are the number of rows and columns. This means df = (5-1)(5-1) = 4*4 =16.
CONCEPT
Pick Your Inference Test
Chi-Square Test for Homogeneity
Chi-Square Test for Association and Independence
10
Edwin conducted a survey to find the percentage of people in an area who smoked regularly. He defined the label “smoking regularly” for males smoking 30 or more cigarettes in a day and for females smoking 20 or more. Out of 635 people who took part in the survey, 71 are labeled as people who smoke regularly.
Edwin wishes to construct a significance test for his data. He finds that the proportion of chain smokers nationally is 14.1%.
What is the z-statistic for this data? Answer choices are rounded to the hundredths place.
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-0.03
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-2.34
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-0.24
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-2.11
RATIONALE
To make things a little easier, let's first note the denominator We can now note that
Finally, subbing all in we find
*note that if you round, the values can be slightly different.
CONCEPT
Z-Test for Population Proportions
11
Select the false statement about ANOVA.
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If a researcher wants to compare the mean wages of females in different age groups at a particular company, he or she should not use an ANOVA because the population means are unknown.
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If a researcher wants to study the effectiveness of three brands of nicotine patches, the researcher should use a One-Way ANOVA test.
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A one-way ANOVA hypothesis test considers comparisons between populations based on
one characteristic, while a two-way ANOVA hypothesis test considers comparisons between populations based on multiple characteristics.
•
If a researcher wants to compare the mean wages of females in different age groups at a particular company to the mean wages of males in different age groups at the same company, the researcher should use a Two-Way ANOVA test.
RATIONALE
If performing a statistical test, we don't need to know the population values. This is true for one-way ANOVA. We use the sample evidence to determine if the means between groups in population are equal.
CONCEPT
One-Way ANOVA/Two-Way ANOVA
12
Rachel measures the lengths of a random sample of 100 screws. The mean length was 2.6 inches, with a standard deviation of 1.0 inches.
Using the alternative hypothesis (µ < µ0), Rachel found that a z-test statistic was equal to -1.25.
What is the p-value of the test statistic? Answer choices are rounded to the thousandths place.
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0.211
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0.317
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0.106
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0.159
RATIONALE
If we go to the chart and the row for the z-column for -1.2 and then the column 0.05, this value corresponds to 0.1056 or 0.106.
CONCEPT
How to Find a P-Value from a Z-Test Statistic
13
The data below shows the heights in inches of 10 students in a class.
The standard error of the sample mean for this set of data is . Answer choices are rounded to the hundredths place.
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1.77
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0.59
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0.19
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1.87
RATIONALE
In order to get the standard error of the mean, we can use the following formula: , 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
14
For a left-tailed test, the critical value of z so that a hypothesis test would reject the null hypothesis at 10% significance level would be . Answer choices are rounded to the hundredths place.
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-1.03
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-2.33
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-1.65
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-1.28 [Show Less]