Question

***This problem must be done using R so please provide the R code used to find the solution. I have provided the data in fitbit.txt below the question. I will also give "thumbs-up for correct R code" Thanks in advance.***

Fitbit, the maker of popular wearable activity trackers, has recently come out with the Charge 4, the successor to the Charge 3. The company claims that the Charge 4 will hold a charge significantly longer than the Charge 3. A writer for a fitness magazine wishes to test the company's claim. She obtains 20 different Charge 3 units and 20 different Charge 4 units. For each unit, she records the time (in days) it takes for the battery to completely drain after a full charge. This data is in the fitbit.txt file. a) Construct a 95% confidence interval for the difference in average time it takes for complete battery drainage using the difference Mcharge4 – Mcharge3. b) Does the interval you constructed in a) suggest that the Charge 4 holds a charge significantly longer than the Charge 3? Please explain why/why not.

Here is the rest of the data needed to solve the problem. Please solve using R. Thank you!

charge3= c(5.66,4.67,5.99,4.54,4.51,5.26,4.59,
5.1,5.46,4.57,4.41,5.58,5.39,5,5.22,
4.5,4.69,5.38,4.6,4.93)

charge4= c(6.05,5.67,7.71,6.46,5.4,5.12,5.76,
4.96,6.22,5.45,4.76,4.45,7.03,5.88,
6.71,6.04,6.32,6.43,5.66,5.94)

Answer 1

The r- codes are provided below along with output, we check the normality condition and it is fullfill so we can use t test.

> charge3=c(5.66,4.67,5.99,4.54,4.51,5.26,4.59,5.1,5.46,4.57,4.41,5.58,5.39,5,5.22,4.5,4.69,5.38,4.6,4.93)
> charge4=c(6.05,5.67,7.71,6.46,5.4,5.12,5.76,4.96,6.22,5.45,4.76,4.45,7.03,5.88,6.71,6.04,6.32,6.43,5.66,5.94)
> shapiro.test(charge3)

   Shapiro-Wilk normality test

data: charge3
W = 0.91872, p-value = 0.09361

> shapiro.test(charge4)

   Shapiro-Wilk normality test

data: charge4
W = 0.98844, p-value = 0.9955

> t.test(charge4,charge3,var.equal=T)

   Two Sample t-test

data: charge4 and charge3
t = 4.4117, df = 38, p-value = 8.178e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.4862048 1.3107952
sample estimates:
mean of x mean of y
   5.9010    5.0025

a) The 95 % confidence interval for mu(charge4)-mu(charge3) is given by,

95 percent confidence interval:
( 0.4862048, 1.3107952)

b) Here the value zero does not lies in the interval, hence we conclude that the charge4 hold a charge significantly longer than the charge3.