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Data Exploration 2. (10 pts) Use to JMP to estimate the 95% confidence interval for the...
C.37 – Grams of Fiber per Day Question C.37 is one of the review exercise questions...
C.37 – Grams of Fiber per
Day
Question C.37 is one of the review exercise questions for Unit C
which is just after Chapter 6.
Use JMP to compute a 95% confidence interval (include output
from JMP, data provided on Canvas): (provided picture)
Interpret interval in the context of this study:
Hypothesis test: ? = 0.05
Hypothesis statements:
Compute test statistic (show calculations). Note that the sample
standard deviation = s = 5.33.
Find p-value using JMP (include output from...
I need help on question 5 and 6 and explaining it. Thank you Annual Salary in...
I need help on question 5 and 6 and explaining it. Thank
you
Annual Salary in 5 Years Quantiles -Summary Statistics Δ. Confidence Intervals H S . . 1000% maximum 150 Mean 74.816667 Parameter Estimate Lower CI Upper CI 1-Alpha 74.81667 70.0473 79.58604 0.950 34.25 Std Err Mean 2.3834993 Std Dev 18.46251 15.64945 22.51802 0.950 18.462506 Mean 99.5% 97.5% 900% 75.0% 500% 250% 100% 2.5% 0.5% 50 Std Dev 100 Upper 95% Mean 79.586038 725 N 57.1 70047296 60 83.75...
2. A study of the thickness of a coating in microns yielded the 25 observations. A...
2. A study of the thickness of a coating in microns yielded the 25 observations. A histogram, stem-and-leaf diagram and boxplot for these data is below: Coating Mean Std Dev Std Err Mean Upper 95% Mean Lower 95% Mean N Stem and Leaf 7.664 0.7330757 0.1466151 7.9665988 7.3614012 25 85 9 9.5 10 Count Stem Leaf 9. 1 Quantiles maximum 9.9 9.9 لبا نے 100.0% 99.5% 97.5% 90.0% 75.0% 50.0% 25.0% 10.0% 2.5% 0.5% 0.0% 3 quartile median quartile 9...
Homes for Sale A real estate agent is interested in the current housing prices in a...
Homes for Sale A real estate agent is interested in the current
housing prices in a particular neighborhood in New York. A random
sample of 30 homes for sale in the neighborhood was taken and the
current listed price of each home was recorded. The data can be
found in the JMP file “HomesForSale-NY-S2019.jmp”
(c) Report the degrees of
freedom associated with the t-distribution required to construct a
confidence interval for the mean for these data.
(d) Based on your...
Small sample difference of means problem. The pathogen Phytophthora capsici causes Bell peppers to wilt and...
Small sample difference of means problem. The pathogen Phytophthora capsici causes Bell peppers to wilt and die. It is thought that too much water aids the spread of the pathogen. Two fields A and B were studied to compare the mean soil water content for the two fields as given by the data below: Field A Field B Summary Statistics Summary Statistics Мean 12.528571 Мean 10.76875 Std Dev Std Dev 2.3940402 2.4010328 Std Err Mean 0.6398342 Std Err Mean 0.6002582...
Quick Check please Respondents Sex Statistic Std. Error 6.33 5.12 7.54 4.85 2.00 90.364 9.506 EmailHrsiperWeek...
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...
Using SPSS DATA, What is the 95% confidence interval for the difference between participant and partner...
Using SPSS DATA, What is the 95% confidence interval for the difference between participant and partner age? A. The partner is from .60 to 1.76 years older than the participant B. The participant is from .60 to 1.76 years older than the partner C. The partner is from 5.88 to 9.45 years older than the participant D. The participant is from 5.88 to 9.45 years older than the partner DATA: Paired Samples Test Paired Differences t df Sig. (2-tailed) Mean...
[1] Use JMP to construct a histogram and box-plot for the variable Receipt Total. [2] Describe the shape of the distribution. Is the distribution roughly symmetric or skewed in a direction? [3] Does t...
[1] Use JMP to construct a histogram and box-plot for the variable Receipt Total. [2] Describe the shape of the distribution. Is the distribution roughly symmetric or skewed in a direction? [3] Does the distribution have any outliers? If so, how many and what are the values? [4] Use JMP to construct a Quantiles table. Paste the quantiles table in the left-hand box and enter the values that make up the five-number summary on the right. maximum 75% quartile median...
4. Use the following charts to compare the central tendency and variability between men and women...
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...
Please write it seriously and clearly 2) Construct 95% Confidence Intervals for the PERCENTAGE o...
Please write it seriously and clearly
2) Construct 95% Confidence Intervals for the PERCENTAGE of
Aluminum ductility and Steel ductility. TREAT THE PERCENTAGE DATA
AS REGULAR DATA VALUES, AND NOT AS PROPORTIONS AS I SUGGESTED
EARLIER. What can you say about the typical proportion ductility
for Aluminum with respect to Steel? From a statistically
significant perspective, can you conclude the Aluminum ductility is
greater than Steel ductility based upon these CIs?
Laboratory 1: Tensile Testing The tensile test can provide...
C.37 – Grams of Fiber per
Day
Question C.37 is one of the review exercise questions for Unit C
which is just after Chapter 6.
Use JMP to compute a 95% confidence interval (include output
from JMP, data provided on Canvas): (provided picture)
Interpret interval in the context of this study:
Hypothesis test: ? = 0.05
Hypothesis statements:
Compute test statistic (show calculations). Note that the sample
standard deviation = s = 5.33.
Find p-value using JMP (include output from...
I need help on question 5 and 6 and explaining it. Thank
you
Annual Salary in 5 Years Quantiles -Summary Statistics Δ. Confidence Intervals H S . . 1000% maximum 150 Mean 74.816667 Parameter Estimate Lower CI Upper CI 1-Alpha 74.81667 70.0473 79.58604 0.950 34.25 Std Err Mean 2.3834993 Std Dev 18.46251 15.64945 22.51802 0.950 18.462506 Mean 99.5% 97.5% 900% 75.0% 500% 250% 100% 2.5% 0.5% 50 Std Dev 100 Upper 95% Mean 79.586038 725 N 57.1 70047296 60 83.75...
2. A study of the thickness of a coating in microns yielded the 25 observations. A histogram, stem-and-leaf diagram and boxplot for these data is below: Coating Mean Std Dev Std Err Mean Upper 95% Mean Lower 95% Mean N Stem and Leaf 7.664 0.7330757 0.1466151 7.9665988 7.3614012 25 85 9 9.5 10 Count Stem Leaf 9. 1 Quantiles maximum 9.9 9.9 لبا نے 100.0% 99.5% 97.5% 90.0% 75.0% 50.0% 25.0% 10.0% 2.5% 0.5% 0.0% 3 quartile median quartile 9...
Homes for Sale A real estate agent is interested in the current
housing prices in a particular neighborhood in New York. A random
sample of 30 homes for sale in the neighborhood was taken and the
current listed price of each home was recorded. The data can be
found in the JMP file “HomesForSale-NY-S2019.jmp”
(c) Report the degrees of
freedom associated with the t-distribution required to construct a
confidence interval for the mean for these data.
(d) Based on your...
Small sample difference of means problem. The pathogen Phytophthora capsici causes Bell peppers to wilt and die. It is thought that too much water aids the spread of the pathogen. Two fields A and B were studied to compare the mean soil water content for the two fields as given by the data below: Field A Field B Summary Statistics Summary Statistics Мean 12.528571 Мean 10.76875 Std Dev Std Dev 2.3940402 2.4010328 Std Err Mean 0.6398342 Std Err Mean 0.6002582...
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...
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...
Please write it seriously and clearly
2) Construct 95% Confidence Intervals for the PERCENTAGE of
Aluminum ductility and Steel ductility. TREAT THE PERCENTAGE DATA
AS REGULAR DATA VALUES, AND NOT AS PROPORTIONS AS I SUGGESTED
EARLIER. What can you say about the typical proportion ductility
for Aluminum with respect to Steel? From a statistically
significant perspective, can you conclude the Aluminum ductility is
greater than Steel ductility based upon these CIs?
Laboratory 1: Tensile Testing The tensile test can provide...
Most questions answered within 3 hours.
-
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