STATISTICS MISC
UNIT 5 – MILESTONE 5
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Joe is measuring the widths of doors he bought to install in an
... [Show More] apartment complex. He measured 72 doors and found a mean width of 36.1 inches with a standard deviation of 0.3 inches. To test if the doors differ significantly from the standard industry width of 36 inches, he computes a z-statistic.
What is the value of Joe's z-test statistic?
2.83
-1.81
-2.83
1.81
RATIONALE
If we first note the denominator of
Then, getting the z-score we can note it is
This tells us that 36.1 is 2.83 standard deviations above the value of 36.
Note that when you round some values you may get slightly different results, but the results should be relatively close to this final calculated value.
CONCEPT
Z-Test for Population Means 2
Emile has calculated a one-tailed z-statistic of -1.97 and wants to see if it is significant at the 5% significance level.
What is the critical value for the 5% significance level? Answer choices are rounded to the hundredths place.
-2.33
0
-1.04
-1.64
RATIONALE
Recall that when a test statistic is smaller than in a left-tailed
test we would reject Ho. The closest value to 5%, or 0.05, in the table would be between 0.0505 and 0.495.
0.0505 corresponds with a z-score of -1.64 0.0495 corresponds with a z-score of -1.65.
We need to calculate the average of the two z-scores to get a z-score of -1.645.
CONCEPT
How to Find a Critical Z Value 3
What do the symbols , , and represent?
Variables of interest
Defined variables
Population parameters
Sample statistics
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 4
A coin is tossed 50 times, and the number of times heads comes up is counted.
Which of the following statements about the distributions of counts and proportions is FALSE?
The distribution of the count of getting heads can be approximated with a normal distribution.
The distribution of the count of getting tails can be approximated with a normal distribution.
The count of getting heads is a binomial distribution.
The count of getting heads from a sample proportion of size 20 can be approximated with a normal distribution.
RATIONALE
If we look at the counts from a large population of success and failures (2 outcomes), this is called a binomial distribution, not a normal distribution.
CONCEPT
Distribution of Sample Proportions 5
Henri has calculated a z-test statistic of -2.73.
What is the p-value of the test statistic? Answer choices are rounded to the thousandths place.
0.003
0.004
0.394
0.006
RATIONALE
If we go to the chart and the row for the z-column for -2.7 and then the column 0.03, this value corresponds to 0.0032 or 0.003.
CONCEPT
How to Find a P-Value from a Z-Test Statistic 6
Keith tabulated the values for the average speeds on each day of his road trip as 61.5, 62.2, 55.7, 50.6, 71.3, 70.8, and 66.8 mph. He wishes to construct a 90% confidence interval.
What value of t* should Keith use to construct the confidence interval? Answer choices are rounded to the thousandths place.
1.943
1.440
RATIONALE
Recall that we have n = 7, so the df = n-1 = 6. So if we go to the
row where df = 7 and then 0.05 for the tail probability, this gives us a value of 1.943. Recall that a 90% confidence interval would have 10% for the tails, so 5% for each tail.
We can also use the last row and find the corresponding confidence level.
CONCEPT
How to Find a Critical T Value 7
The data below shows the daily low temperatures, in degrees Fahrenheit, of a city for one week.
Day Low Temperature, in Fahrenheit
Monday 54.5
Tuesday 53
Wednesday 56.5
Thursday 54
Friday 52.5
Saturday 51
Sunday 53
The standard error of the sample mean for this set of data is . Answer choices are rounded to the hundredths place.
0.65
0.25
1.73
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 1.73. The sample size is seven days.
The standard error is then:
CONCEPT
Calculating Standard Error of a Sample Mean 8
A table represents the number of students who passed or failed an aptitude test at two different campuses.
South Campus North Campus Passed 42 31
Failed 58 69
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?
36.5 students
42 students
43.7 students
50 students
RATIONALE
In order to get the expected counts we can note the formula is:
CONCEPT
Chi-Square Test for Homogeneity 9
A researcher has a table of data with 5 column variables and 4 row variables.
The value for the degrees of freedom in order to calculate the statistic is .
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20
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