male
first quartile is the mid value between minimum value and median
minimum =0, median = 2
therefore first quartile = (2-0)/2= 1
third quartile is the mid value between median and maximum value
median = 2, maximum = 50
therefore, third quartile = (50-2)/2= 24
female
first qurtile = 1
third quartile = 24
Respondents Sex Statistic Std. Error 6.33 5.12 7.54 4.85 2.00 90.364 9.506 EmailHrsiperWeek Male Mean 95%...
Respondents Sex Statistic Std. Error 6.33 5.12 7.54 4.85 2.00 90.364 9.506 EmailHrsiperWeek Male Mean 95% Confidence Interval for Mean 5% Trimmed Mean Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis 615 Lower Bound Upper Bound 50 50 2.658 7.585 5.93 4.98 6.88 4.57 2.00 78.924 8.884 157 314 485 Female Mean 95% Confidence Interval for Mean 5% Trimmed Mean Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis Lower Bound Upper Bound 50 50...
Respondents Sex Statistic Std. Error 6.33 5.12 7.54 4.85 2.00 90.364 9.506 EmailHrsiperWeek Male Mean 95% Confidence Interval for Mean 5% Trimmed Mean Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis 615 Lower Bound Upper Bound 50 50 2.658 7.585 5.93 4.98 6.88 4.57 2.00 78.924 8.884 157 314 485 Female Mean 95% Confidence Interval for Mean 5% Trimmed Mean Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis Lower Bound Upper Bound 50 50...
Quick Check please
Respondents Sex Statistic Std. Error 6.33 5.12 7.54 4.85 2.00 90.364 9.506 EmailHrsiperWeek Male Mean 95% Confidence Interval for Mean 5% Trimmed Mean Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis 615 Lower Bound Upper Bound 50 50 2.658 7.585 5.93 4.98 6.88 4.57 2.00 78.924 8.884 157 314 485 Female Mean 95% Confidence Interval for Mean 5% Trimmed Mean Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis Lower Bound Upper...
What does the five number tell us about the time spent on email
(Hint, interpret the five number summary in plain English) and what
does the Boxplot and the normality test show? Explain.
Use the 1.5xIQR rule to identify possible outliers. List the
cutoff points for outliers, Show your workings. Explain what you
found out. (Hint: Are there any excessive time spent on email for
Male(1) or Female(2) or both).
GET DATA /TYPE-XLS /FILE='C: \Users\rmanda 1 \ Desktop\homework! . xls'...
4. Use the following charts to compare the central tendency and variability between men and women for the variable AGEKDBRN and answer the following questions. Case Processing Summary Cases Missing Valid Total sex RESPONDENTS SEX 1 MALE 2 FEMALE Percent 601 | 67.5% 869| 75.4% Percent 891 | 100.0% 284 | 24.6% | 1153 | 100.0% Percent 2901 32.5% WHEN 1ST CHILD BORN Descriptives StatisticStd. Error 269 agekdbrn RS AGE WHEN 1ST CHILD BORN 25.54 25.02 Upper Bound 26.07 25.13...