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The introductory statement in the question indicates: The data set to
The introductory statement in the question indicates: The data set to
"Total family income" [income98] is quantitative (ordinal treated as
"Total family income" [income98] is quantitative (ordinal treated as
To evaluate the variables conformity to the nearly normal condition,
To evaluate the variables conformity to the nearly normal condition,
Move the variable for the analysis income98 to the Variable(s) list
Move the variable for the analysis income98 to the Variable(s) list
The check boxes for Mean and Std
The check boxes for Mean and Std
Click on the OK button to produce the output
Click on the OK button to produce the output
"Total family income" [income98] satisfied the criteria for a normal
"Total family income" [income98] satisfied the criteria for a normal
Sort the column Zincome98 in ascending order to show any negative
Sort the column Zincome98 in ascending order to show any negative
Sort the column Zincome98 in descending order to show any positive
Sort the column Zincome98 in descending order to show any positive
"Total family income" [income98] satisfied the criteria for a normal
"Total family income" [income98] satisfied the criteria for a normal
The number of valid cases available for the test was 229, larger than
The number of valid cases available for the test was 229, larger than
To produce the one-sample t-test of a population mean, select the
To produce the one-sample t-test of a population mean, select the
Third, click on the OK button to produce the output
Third, click on the OK button to produce the output
The values we need are in the One-Sample Statistics table
The values we need are in the One-Sample Statistics table
The correct value for the sample mean was 15
The correct value for the sample mean was 15
The null hypothesis for a one-sample t-test of a population mean
The null hypothesis for a one-sample t-test of a population mean
The statement is correct and contains the correct values for both the
The statement is correct and contains the correct values for both the
The probability that a sample with a mean of 15
The probability that a sample with a mean of 15
The correct p-value for this test (p =
The correct p-value for this test (p =
Since we rejected the null hypothesis and since the mean of "total
Since we rejected the null hypothesis and since the mean of "total
To use the T-Tests script, select the type of T-Test desired by
To use the T-Tests script, select the type of T-Test desired by
Select a test variable from the list
Select a test variable from the list
The script produces skewness and kurtosis as descriptive statistics
The script produces skewness and kurtosis as descriptive statistics
The T-Test output shows the number of cases for evaluating the
The T-Test output shows the number of cases for evaluating the
The script also produces a histogram and boxplot
The script also produces a histogram and boxplot
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SOLVING THE PROBLEM A one-sample t-test of a population mean requires that the variable be quantitative

содержание презентации «SOLVING THE PROBLEM A one-sample t-test of a population mean requires that the variable be quantitative.ppt»
Сл Текст Сл Текст
1SOLVING THE PROBLEM A one-sample 20Third, click on the OK button to
t-test of a population mean requires that produce the output. First, move the
the variable be quantitative. A one-sample variable income98 to the Test Variable(s)
test of a population mean tests the null list box. Second, type the value of the
hypothesis that the population mean based population mean that we are testing
on our sample data is not really different against, i.e. the mean based on previous
from some value specified for the research (16.5). 8/7/2015. Slide 20.
population mean, e.g. identified in 21The values we need are in the
previous research. The hypothesis is One-Sample Statistics table. The correct
evaluated using probabilities for a value for the sample mean was 15.7. The
sampling distribution that is normal (the standard error of the sampling
t distribution). This requires the distribution was correctly identified as
determination that our data supports the 0.349. 8/7/2015. Slide 21.
use of a normal distribution. We satisfy 22The correct value for the sample mean
this requirement either by demonstrating was 15.7. The standard error of the
that our sample data which is sampling distribution was correctly
representative of the population is identified as 0.349. Mark the check box as
reasonably normal (symmetric, unimodal, correct. 8/7/2015. Slide 22.
and without extreme outliers), or by 23The next statement asks us about the
demonstrating that our sample is null hypothesis for the one-sample t-test.
sufficiently large that the sampling We should check to make certain the
distribution from which our sample comes relationship is stated correctly, and that
will follow a normal distribution even we verify the correct for the population
when though our individual sample does not mean that we are testing against.
(the Central Limit Theorem). If our sample 8/7/2015. Slide 23.
mean is close to the specified population 24The null hypothesis for a one-sample
mean (using the standard error of the t-test of a population mean states that
distribution as the measure of closeness), the estimated population mean of income98
the probability that our sample could be based on our data (15.7) is not different
drawn from a population with the specified from the population mean of income98 based
population mean will be high, and we fail on previous research (16.5), i.e., the two
to reject the null hypothesis and do not estimates of the true population mean are
interpret the results. 8/7/2015. Slide 1. equivalent. Mark the statement as correct.
2If our sample mean is far away from 8/7/2015. Slide 24.
the specified population mean (using the 25The next statement asks us to relate
standard error of the distribution as the the t-test to the data in our problem.
measure of distance), the probability that 8/7/2015. Slide 25.
our sample could be drawn from a 26The t-test statistic is based on the
population with the specified population difference between the estimated
mean will be low. We reject the null population mean based on our data and the
hypothesis and interpret the results of hypothesized population mean based on
the test, consistent with the alternative previous research (15.668 - 16.500 =
hypothesis that explains the difference -0.832) relative to the standard error of
between the population mean based on our the sampling distribution (0.349).
sample and the population mean specified 8/7/2015. Slide 26.
for the test. The decision that our sample 27The statement is correct and contains
mean is close or distant from the the correct values for both the difference
specified population mean is based on a in means and the sampling error that we
comparison of the probability of the would typically expect to find in the
t-test statistic to the alpha level of sampling distribution for income98. Mark
significance established for the problem. the statement as correct. 8/7/2015. Slide
8/7/2015. Slide 2. 27.
3The introductory statement in the 28The next statement asks about the
question indicates: The data set to use probability for the comparison made by the
(GSS2000R) The variable to use in the t-test. i.e. what is the probability that
analysis: total family income [income98] sample mean with the value for our data
The task to accomplish (one-sample t-test could be drawn from a population with the
of a population mean) The population mean mean based on previous research and the
(16.5) The level of significance (0.05, standard error estimated from our data.
two-tailed). 8/7/2015. Slide 3. 8/7/2015. Slide 28.
4The first statement asks about the 29The probability that a sample with a
level of measurement. 8/7/2015. Slide 4. mean of 15.7 could be drawn from a
5"Total family income" population with a mean of 16.5 was p =
[income98] is quantitative (ordinal .018, not p = .038. 8/7/2015. Slide 29.
treated as quantitative), satisfying the 30The probability that a sample with a
level of measurement requirement for a mean of 15.7 could be drawn from a
one-sample t-test of a population mean. population with a mean of 16.5 was p =
8/7/2015. Slide 5. .018, not p = .038. The statement is not
6To justify the use of probabilities correct and the check box is left empty.
based on a normal sampling distribution in 8/7/2015. Slide 30.
testing hypotheses, either the 31When the p-value for the statistical
distribution of the variable must satisfy test is less than or equal to alpha, we
the nearly normal condition or the size of reject the null hypothesis and interpret
the sample must be sufficiently large to the results of the test. If the p-value is
generate a normal sampling distribution greater than alpha, we fail to reject the
under the Central Limit Theorem. A null hypothesis and do not interpret the
one-sample t-test of a population mean result. 8/7/2015. Slide 31.
requires that the distribution of the 32The correct p-value for this test (p =
variable satisfy the nearly normal .018) is less than or equal to the alpha
condition, which we will operationally level of significance (p = .050)
define as having skewness and kurtosis supporting the conclusion to reject the
between -1.0 and +1.0, and having no null hypothesis. Mark the statement as
outliers with standard scores equal to or correct. 8/7/2015. Slide 32.
smaller than -3.0 or equal to or larger 33The final statement asks us to
than +3.0. 8/7/2015. Slide 6. interpret the result of our statistical
7To evaluate the variables conformity test as a finding in the context of the
to the nearly normal condition, we will problem we created. We only interpret the
use descriptive statistics and standard results when the null hypothesis is
scores. To compute the descriptive rejected. 8/7/2015. Slide 33.
statistics and standard scores, select the 348/7/2015. Slide 34.
Descriptive Statistics > Descriptives 35Since we rejected the null hypothesis
command from the Analyze menu. 8/7/2015. and since the mean of "total family
Slide 7. income" in our sample (15.7) is
8Move the variable for the analysis actually smaller than the mean reported in
income98 to the Variable(s) list box. previous research (16.5), it is reasonable
Click on the Options button to select to suggest that true mean of "total
optional statistics. 8/7/2015. Slide 8. family income" in the population has
9The check boxes for Mean and Std. decreased. Mark the check box as correct.
Deviation are already marked by default. 8/7/2015. Slide 35.
Click on Continue button to close the 36Variable is quantitative? Do not mark
dialog box. Mark the Kurtosis and Skewness check box. No. Mark only “None of the
check boxes. This will provide the above.”. Stop. Mark statement check box.
statistics for assessing normality. 8/7/2015.
8/7/2015. Slide 9. 37Variable nearly normal distribution?
10Click on the OK button to produce the Nearly normal: Skewness between -1.0 and
output. Mark the check box Save +1.0 Kurtosis between -1.0 and +1.0
standardized values as variables. Z-scores between -3.0 and +3.0. Do not
8/7/2015. Slide 10. mark check box. No. CLT stands for Central
11"Total family income" Limit Theorem. CLT applicable (Sample size
[income98] satisfied the criteria for a ? 40)? Mark statement check box. Do not
normal distribution. . The skewness of the mark check box. CLT applicable (Sample
distribution (-0.628) was between -1.0 and size ? 40)? Stop. If the difference
+1.0. The kurtosis of the distribution variable is not normal and the sample size
(-0.248) was between -1.0 and +1.0. is less than 40, the test is not
8/7/2015. Slide 11. appropriate. Do not mark check box. Mark
12Sort the column Zincome98 in ascending statement check box. The path is
order to show any negative outliers at the complicated because we check two
top of the column. There were no outliers conditions, but only one needs to be
that had a standard score less than or correct to continue. 8/7/2015. Slide 37.
equal to -3.0. 8/7/2015. Slide 12. 38Sample mean and standard error
13Sort the column Zincome98 in correct? No. Do not mark check box. Mark
descending order to show any positive statement check box. H0: mean = mean from
outliers at the top of the column. There previous research. Do not mark check box.
were no outliers that had a standard score No. Mark statement check box. 8/7/2015.
greater than or equal to +3.0. 8/7/2015. Slide 38.
Slide 13. 39T-test accurately described? Do not
14"Total family income" mark check box. No. Mark statement check
[income98] satisfied the criteria for a box. P-value (sig.) stated correctly? Do
normal distribution. The skewness of the not mark check box. No. Mark statement
distribution (-0.628) was between -1.0 and check box. 8/7/2015. Slide 39.
+1.0 and the kurtosis of the distribution 40Reject H0 is correct decision (p ?
(-0.248) was between -1.0 and +1.0. There alpha)? Do not mark check box. No. Stop.
were no outliers that had a standard score We interpret results only if we reject
less than or equal to -3.0 or greater than null hypothesis. Mark statement check box.
or equal to +3.0. Mark the statement as Interpretation is stated correctly? Do not
correct. 8/7/2015. Slide 14. mark check box. No. Mark statement check
15Though we have satisfied the nearly box. 8/7/2015. Slide 40.
normal condition and do not need to 41To use the T-Tests script, select the
utilize the Central Limit Theorem to type of T-Test desired by marking one of
justify the use of probabilities based on the option buttons. 8/7/2015. Slide 41.
the normal distribution, we will still 42Select a test variable from the list.
examine the sample size. To apply the Specify the value for the population mean
Central Limit Theorem for a one-sample to test against. Click on the OK button to
t-test of a population mean requires that produce the output. 8/7/2015. Slide 42.
the sample have 40 or more cases. 43The script produces skewness and
8/7/2015. Slide 15. kurtosis as descriptive statistics that
16The number of valid cases available can be used to evaluate the near normal
for the test was 229, larger than condition. The script also creates a table
requirement of 40 cases to apply the of outliers that will list any outliers
Central Limit Theorem. However, since the with a z-score less than or equal to -3.0
distribution of "total family or greater than or equal to + 3.0. If
income" satisfied the nearly normal there are no outliers, it will list the
condition, we do not need to make use of minimum and maximum z-score for the data.
the Central Limit Theorem to satisfy the 8/7/2015. Slide 43.
sampling distribution requirements of a 44The T-Test output shows the number of
one-sample t-test of a population mean. cases for evaluating the applicability of
8/7/2015. Slide 16. the Central Limit Theorem. The T-Test
17The number of valid cases available output provides the evidence needed to
for the test was 229, larger than make a decision about the null hypothesis.
requirement of 40 cases to apply the 8/7/2015. Slide 44.
Central Limit Theorem. Mark the statement 45The script also produces a histogram
as correct. 8/7/2015. Slide 17. and boxplot. that can be used to evaluate
18The next statement asks us to identify the distribution or confirm your
the mean for the sample data and the interpretation of the statistical
standard error of the sampling evidence. To facilitate interpretation,
distribution. To answer this question, we the histogram can be overlaid with a
need to produce the output for the dotplot showing more detail for the
one-sample t-test. 8/7/2015. Slide 18. distribution of cases, lines showing the
19To produce the one-sample t-test of a location of the mean and standard
population mean, select the Compare Means deviation units, and a normal curve.
> One-Sample T Test command from the 8/7/2015. Slide 45.
Analyze menu. 8/7/2015. Slide 19.
SOLVING THE PROBLEM A one-sample t-test of a population mean requires that the variable be quantitative.ppt
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Лондон

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900igr.net > Презентации по английскому языку > Лондон > SOLVING THE PROBLEM A one-sample t-test of a population mean requires that the variable be quantitative