Question

Case study. Please provide a step by step explanation. Thank you!

CASE STUDY Fire-Safe Cigarettes U.90,000 fires in the United States were started Naturally, cigarette manufacturers were concerned s sighted tobacco product. The dollar value of the about the additional cost of manufacturing such a lost in these fires is staggering, In that same year. cigarette. Plas, the man afacturers were concerned that ins caused by dropped or dicarded cigarettes resulted in consumers might prefer the cigarettes without "speed daths Twenty-Eive percent of the victims of cigarette bumps" to those with design changes. One particular yWed fires were not the smokers and about 40% of measure the manufacturers were concerned about was aarcme-induced fires started in the bedroom. (Source the amount of nicotine in each cigarette. The following data represent the amount of nicotine in each brand of Certainly, it makes sense to develop a cigarctte that ciganctle for a random sampke of 15 cigaretties cinguishes iself when left unattended (as would happen twcone falls asleep while smoking). The state of New ork decided to require cigarettes sold in its stores to be va types od cigarettes were tested to determine their opessity to self-estinguish. For each type (A and B).40 iproties scre lt and allowed to burn unattended until the aprete either extinguished or did not. Cigarette brand A 134 113 designed to have ultra thin concentric paper bands sod to the traditional cigarette paper. These bands are nered to as "speed bumps" and cause extinguishing d the cigarette by restricting the fow of oxygen to the urning cmber. Cigarette brand B was a traditionally eigned model The resalts of the experiment are in Write a report detailing the propensity of each brand to self-extinguish and the level of nicotine in each brand. Include any relevant confidence intervals. If the normal model or Student's -distribution caneot be used to construct a coafidence interval, use alternative models (sech as the Agresti-Coull model for estimating proportions-sce Problcms 45 and 46 from Section 9.1 or the bootstrap methods of Section 9.5). Would yos support legislation that redaced the risk of tire from unatended cigarettes? How mach extra wowld you be willing to pay lowing table, where E represents an estinguished e and F represents a full-barn cigarette. foe sach a cigarette if you scre a smoker?

Answer 1

Ans. Since 40 sample data of brand A & brand B are categorical & we need to find the proportion of Extinguished cigarette & did not extinguish cigarette , so we will do 1 sample proportion test on minitab for both of the brand.

Look at below for arrangement of my data :-

Br E E E E E E F F F F E E F E E E E E E E E E E E E F E F F E F E E F F F EF 5 m m ㅡ 0 | 1 2 3 4 S 6 7 9 0 1 2 3 4 5

Now I perform a one-sample proportion test to find the propensity of each brand to be extinguished.

Look at the result of minitab below for Brand A

Test and CI for One Proportion: BrandA Method Event: BrandA F p: proportion where Brands F Exact method is used for this analysis. Descriptive Statistics N Event Sample p 95% CI for p 40 13 0.325000 (0.185729, 0.491295) Test Null hypothesis Alternative hypothesis H1:p # 0.5 P-Value Ho: p 0.5 0.038

In the above result event of brand A is F , that is did not extinguish.

p = proportion that brand A = F

Now I take null hypothesis that H0 : There is equal no. of E & F cases in Brand A.

Ha : Proportion is not equal

So, H0 : p = 0.05

Ha : p 0.05

But our result shows that out of 40 cases 13 are showing F , so 32.5% cases are F. So we can reject our null that 50% of Brand A did not extinguished when it is unattended. We could make our decision by seeing the p-value also, which is p = 0.038 < 0.05 in this case.

Similar process has been done for Brand B.

Please look at the below image :-

Test and CI for One Proportion: BrandB Method Event BrandB F p: proportion where Brande F Exact method is used for this analysis. Descriptive Statistics N Event Sample p 95% Cl for p 40 39 0.975000 (0.868414, 0.999367) Test Null hypothesis Alternative hypothesis H1:p # 0.5 P-Value Ho: p = 0.5 0.000

Here out of 40 observation 39 are F. So here proportion of is greater than what we have assumed. So here we also reject null hypothesis. 97.5% observation did not extinguished fire. & p-value is 0. So probability is 0 for null to be true.

....... Now the amount of Nicotine in each brand ......

Now if the data of Nicotine across two brands follows normal distribution , then we can test 2-sample t-test to find that whether there is any difference of the mean of the nicotine across two brands.

At first see the dataset below :-

Brand Nicotine 1.4 1.36 1.42 1.3 1.4 1.09 1.06 1.06 1.12 1.12 1.14 1.14 1.13 1.21 1.15 1.24 1.36 1.32 1.32 1.36 1.23 1.13 1.2 1.28 1 A 2 A 3 A 4 A 2 B 5 B

Now with this data at first we do normality test & HOV test. Result of normality test u can show below...U can do it on SPSS easily

By seeing the third test of normality table where sig. column of shapiro wilk test show the p value of accepting the null hypothesis that is the data is normally distributed. P = .075 > .05. Also population mean of Brand A & Brand B is 1.2203 (from 2nd descriptives table).

er Case Processing Summary Cases Valid Missing Total Percent Percent Percent set essi Cr Nicotine 30 | 1 00.0% 0.0% 30 | 1 00.0% Descriptives Tes Statistic Std. Error Nicotine Mean 1.2203 Lower Bound1802 Upper Bound 1.2604 1.2185 1.2000 012 10743 1.06 1.42 36 01961 95% Confidence Interval for Mean 5% Trimmed Mean Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis set er set essi rm 394 1.002 427 al de 833 Tests of Normality Kolmogorov-Smirnova df Shapiro-Wilk df Statistic 142 Statistic 937 Sig Sig set Nicotine 30 128 30 075 a. Lilliefors Significance Correction

Now the 2-sample t-test

Null hypothesis :- H0 : There is no difference in the amount of nicotine across two brands.

Ha : There is a mean difference of nicotine.

Please find the result of t-test on SPSS,

T-Test [DataSet3] Group Statistics Std. Error Mean Mean Std. Deviation 13151 07881 Brand Nicotine A 15 1.2067 15 1.2340 03396 02035 Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means 95% Confidence Interval of the Difference Mean Std. Error Difference Sig df Sig. (2-tailed Difference Lower Upper Nicotine Equal variances 6.871 014 .690 28 496 .02733 03959 .10842 05376 assumed Equal variances not assumed -,690 22.907 497 ,02733 03959 10924 05458

Now from above image independent sample test table shows that HOV condition is not satisfied. That is there is difference in the variances of nicotine across two brands (p = 0.014 < 0.05). But here we can accept our null hypothesis by seeing again the same table & looking at the p value of equal variances not assumed case , which shows p = 0.497 > 0.05. So , there is not any difference in the amount of nicotine across two brands. Also mean values of the sample Brand A & Brand B are 1.2067 & 1.2340 respectively. which is showing a negligible difference.

Lastly,the legislation that reduced the risk from fire of unattended cigarettes might be downward in case of Brand A but not the same in case of Brand B.

Because proportion test shows that in case of Brand A 32.5% of cigrettes that did not extinguished fire but Brand B shows 97.5% undistinguished case.