## We use Minitab 16 Software to solve the problems.

Question #1.

– Find the range, variance and standard deviation of this data set

## Descriptive Statistics: NICOTINE

Variable N N* Mean SE Mean StDev Variance Minimum Q1 Median

NICOTINE 500 0 0, 8425 0, 0155 0, 3455 0, 1194 0, 0500 0, 7000 0, 9000

Variable Q3 Maximum Range

NICOTINE 1, 1000 1, 9000 1, 8500

– Create the relative frequency histogram for the nicotine level.

## Graph -> Histogram -> With Fit

– Compute the interval y ± 2s

y = 0. 8425s= 0. 3455y +2s= 0. 8425+2*0. 3455= 1. 5335

y -2s= 0. 8425-2*0. 3455= 0. 1515

Hence, the interval has the following form: (0. 1515; 1. 5335)

– Estimate the percentage of brands with nicotine levels less than 0. 5 mg

## There are 62 observations with nicotine levels less than 0. 5 mg out of 500 total observations

62500= 0. 124= 12. 4%

## Question #2

– Construct the Normality plots for Trial 4 and Trial 5.

## Graph -> Probability Plot -> Single

– Construct comparative box plots of these measurements.

## Graph -> Box Plot -> Multiple Y’s (Simple)

– Does it seem that all five trials are consistent with respect to the variability of the measurements?

No, there is significant difference between variability of the measurements for all five trials. The range and measures of central tendency values are significantly different.

– Are all five trial centered on the same value?

## No, each trial centered on different values, we can see it from each box plot (horizontal line within the box)

– How does each group of trials compare to the true value?

## On average, the true value is lower than values in each of five trials.

– Could there have been bias in the measuring instrument?

Of course it possible, we have measured all observations with subtracting of 299, 000. This procedure may cause bias estimations.

## Question #3

– Describe the MPG Mileage Ratings of the 100 cars with a Boxplot. Show the outliers present in the data along with values of the corresponding outliers.

## Graph -> Box Plot -> One Y (Simple)

There are two outliers marked with *. These values are 44. 9 and 30

– Display a table of the descriptive statistics (i. e. variance, mean, squared error, standard deviation, Q1, Q3 and IQR)

## Stat -> Basic Statistics –> Display Descriptive Statistics

Descriptive Statistics: MPG

Variable N N* Mean SE Mean StDev Variance Minimum Q1 Median

MPG 100 0 36, 994 0, 242 2, 418 5, 846 30, 000 35, 625 37, 000

Variable Q3 Maximum Range IQR

MPG 38, 375 44, 900 14, 900 2, 750

Question #4

– Draw and compare the dot diagram for the two data sets

## Graph -> Dot Plot -> Multiple Y’s (Simple)

– Draw Box Plots to assist in the interpretation of the tension strength data from this experiment.

Graph -> Box Plot -> Multiple Y’s (Simple)

– Are there any outliers in the data?

## No, it seems that there are no outliers in both data sets.

Question #5

– Is there evidence to support the assumption that foam height is normally distributed?

Graph -> Probability Plot -> Single

The Anderson-Darling test shows p-value at a level of 0. 504. Hence, the data cannot be considered as normally distributed data.

– Find a 95% CI on the mean foam height

## Stat -> Basic Statistics -> 1-Sample t

One-Sample T: C1

Variable N Mean StDev SE Mean 95% CI

C1 10 203, 20 7, 50 2, 37 (197, 84; 208, 56)

– Find a 95% prediction interval on the next bottle of shampoo that will be tested

## Stat -> Basic Statistics -> 1-Sample t

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI

1 199, 00 5, 17 (187, 09; 210, 91) (177, 88; 220, 12)

## Values of Predictors for New Observations

New Obs C2

1 11, 0

– Find an interval that contains 95% of the shampoo foam heights with 99% confidence.

– Explain the difference in the intervals computed in parts (b), (c) and (d).

## The confidence interval is interval in which the true population values are within .

The prediction interval is interval in which the predicted values (results of estimations) are within.

Question #6

– Construct a summary table and a horizontal bar graph to describe the ice types of the 504 melt ponds.

## Stat -> Basic Statistics – Graph Summary

– Find a 90% confidence interval for the true mean visible albedo value of all Canadian Arctic ice ponds in column L to T.

## Stat -> Basic Statistics -> 1-Sample t

One-Sample T: broadband-al; visible-alb; infrared-alb; band1-alb;

Variable N Mean StDev SE Mean 90% CI

broadband-alb 504 0, 21028 0, 09802 0, 00437 (0, 20308; 0, 21747)

visible-alb 504 0, 36024 0, 13561 0, 00604 (0, 35028; 0, 37019)

infrared-alb 504 0, 05990 0, 07829 0, 00349 (0, 05415; 0, 06565)

band1-alb 504 0, 44107 0, 16138 0, 00719 (0, 42923; 0, 45292)

band2-alb 504 0, 42595 0, 15717 0, 00700 (0, 41442; 0, 43749)

band3-alb 504 0, 25873 0, 13293 0, 00592 (0, 24897; 0, 26849)

band4-alb 504 0, 08258 0, 09852 0, 00439 (0, 07535; 0, 08981)

band5-alb 504 0, 06798 0, 08125 0, 00362 (0, 06201; 0, 07394)

band6-alb 504 0, 05270 0, 07696 0, 00343 (0, 04705; 0, 05835)

– Find a 90% confidence interval for the true variance visible albedo value of all Canadian Arctic ice ponds in column L to T.

## Stat –> Basic Statistics -> 1 Variance

Test and CI for One Variance: broadband-al; visible-alb; infrared-alb;

Method

The chi-square method is only for the normal distribution.

The Bonett method is for any continuous distribution.

Statistics

Variable N StDev Variance

broadband-alb 504 0, 0980 0, 00961

visible-alb 504 0, 136 0, 0184

infrared-alb 504 0, 0783 0, 00613

band1-alb 504 0, 161 0, 0260

band2-alb 504 0, 157 0, 0247

band3-alb 504 0, 133 0, 0177

band4-alb 504 0, 0985 0, 00971

band5-alb 504 0, 0813 0, 00660

band6-alb 504 0, 0770 0, 00592

## 90% Confidence Intervals

Variable Method CI for StDev CI for Variance

broadband-alb Chi-Square (0, 0932; 0, 1034) (0, 00869; 0, 01069)

## Bonett (0, 0903; 0, 1067) (0, 00816; 0, 01139)

visible-alb Chi-Square ( 0, 129; 0, 143) ( 0, 0166; 0, 0205)

## Bonett ( 0, 129; 0, 143) ( 0, 0167; 0, 0204)

infrared-alb Chi-Square (0, 0744; 0, 0826) (0, 00554; 0, 00682)

## Bonett (0, 0667; 0, 0922) (0, 00445; 0, 00850)

band1-alb Chi-Square ( 0, 153; 0, 170) ( 0, 0235; 0, 0290)

## Bonett ( 0, 154; 0, 170) ( 0, 0237; 0, 0288)

band2-alb Chi-Square ( 0, 149; 0, 166) ( 0, 0223; 0, 0275)

## Bonett ( 0, 150; 0, 165) ( 0, 0225; 0, 0274)

band3-alb Chi-Square ( 0, 126; 0, 140) ( 0, 0160; 0, 0197)

## Bonett ( 0, 124; 0, 143) ( 0, 0154; 0, 0204)

band4-alb Chi-Square (0, 0937; 0, 1039) (0, 00878; 0, 01080)

## Bonett (0, 0857; 0, 1136) (0, 00735; 0, 01291)

band5-alb Chi-Square (0, 0773; 0, 0857) (0, 00597; 0, 00735)

## Bonett (0, 0705; 0, 0940) (0, 00497; 0, 00883)

band6-alb Chi-Square (0, 0732; 0, 0812) (0, 00536; 0, 00659)

## Bonett (0, 0674; 0, 0881) (0, 00455; 0, 00776)

– What is the interpretation of the intervals gotten in (b) and (c)?

These intervals shows that we are 95 confident that true mean visible albedo value and true variance visible albedo value are in those intervals (for each type from column L to T).