Researchers would like to know whether there is a significant difference between
ID: 3206726 • Letter: R
Question
Researchers would like to know whether there is a significant difference between the average battery life of laptops with a CD writer and those without. To understand the effect of CD writing on laptop batteries, 23 users without a CD writer and 17 users with a CD writer are asked to work on their laptops until the “low battery” sign came on. The following software output is an analysis of this data:
Std. Dev.
n
No CD Writer
5.09
0.64
23
CD Writer
4.67
0.72
17
Do these data provide evidence of a difference between the average battery life of laptops with a CD writer and those without? Assume data is obtained randomly. Use a significance level of 0.10 and answer the following questions using the software output.
a)Is there visual evidence that the average battery life is different among the laptops with and without the CD writer? Explain.
b)State the null and alternative hypothesis to answer the question of interest. Is the test one or two sided?
c)Check conditions for the test. State each and whether it is met.
d)Using the R output give a four part conclusion and thoroughly answer the question of interest.
R code
# This output is already provided to you in the data analysis but is included here
> # for an optional look at how a two sample t test can be done in R.
> # CD WRITER DATA
> writer = c(rep("CD-Writer", 17), rep("No CD-Writer", 23))
> batterytime =c(3.69, 4.45, 5.33, 4.10, 4.94, 4.27, 4.96, 3.60, 5.23, 3.94,
+ 4.20, 5.70, 5.06, 3.66, 5.26, 5.64, 5.41, 5.64, 4.64, 4.29,
+ 6.24, 4.99, 3.95, 5.36, 5.57, 6.28, 5.17, 4.81, 6.00, 4.54,
+ 4.43, 4.87, 4.74, 5.01, 6.06, 4.93, 5.28, 4.43, 4.76, 5.14)
> CDdata = as.data.frame(cbind(as.factor(writer),batterytime))
> aggregate(batterytime~writer, CDdata, mean)
writer batterytime
1 CD-Writer 4.672941
2 No CD-Writer 5.092609
> aggregate(batterytime~writer, CDdata, sd)
writer batterytime
1 CD-Writer 0.7231940
2 No CD-Writer 0.6351179
> boxplot(batterytime~writer,
+ main = "Possible Effect of CD Writers on Laptop Batteries",
+ xlab = "Battery Time in Hours",
+ horizontal = TRUE, col = c("tomato1", "royalblue2"))
> t.test(batterytime~writer, conf.level = 0.90)
Welch Two Sample t-test
data: batterytime by writer
t = -1.9095, df = 31.902, p-value = 0.06523
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
-0.79198476 -0.04735027
sample estimates:
mean in group CD-Writer mean in group No CD-Writer
4.672941 5.092609
> boxplot(sunspots$spot, horizontal = TRUE, main = "Add a title here ;) ", col = "firebrick2", xlab = "Add an x title here")
Mean
Std. Dev.
n
No CD Writer
5.09
0.64
23
CD Writer
4.67
0.72
17
Possible Effect of CD Writers on Laptop Batteries 6.0 4.0 4.5 5.0 3.5 5.5 Battery Time in HoursExplanation / Answer
Solution
Let X = battery life of laptops with No CD writer and Y = battery life of laptops with CD writer
Assumption: X ~ N(µ1, 12) and Y ~ N(µ2, 22).
Since 12 and 22 are not known, a pre-requisite for applying standard t-test for testing µ1 = µ2 is to test
12 = 22. So, we first test(at 5% level of significance, the most frequently used level)
H0: 12 = 22 Vs HA: 12 22. For this, the Test Statistic, F = s22/s12 , where s12 and s22 are respective sample variances for ‘No CD writer ‘ and ‘with CD writer’.[for F, larger variance is always at the numerator.]
Under H0, F has F-distribution with degrees of freedom, (n2 - 1) and (n1 - 1), where n1 and n2 are the respective sample sizes.
Now given, n1 = 23. n2 = 17, s12 = 0.642 and s22 = 0.722, F = 1.2656
From Standard Statistical Tables, upper 5% point of F-distribution with degrees of freedom, 16 and 22, is 2.24 which is more than the calculated value 1.2656 of F. Hence, H0 accepted.
=> 12 = 22 = 2 , say.
Then, 2 can be estimated by s2 = {(n1 – 1)s12 + (n2 – 1)s22)/(n1 + n2 - 2) = 0.4554 and hence
s = 0.6748.
Now, to test whether there is significant difference between the average life of the two groups of laptops,
H0: µ1 = µ2 Vs HA: µ1 µ2. For this, the Test Statistic, t = |(x bar – y bar)|)/[s. .{(1/n1) + (1/n2)}] has, Under H0, t-distribution with degrees of freedom, (n1 + n2 - 2), where x bar and y bar are the respective sample means.
So, t = 0.42/{0.6748.(0.0435 + 0.0588)} = 0.42/0.2158 = 1.94.
From Standard Statistical Tables, upper 5% point of t-distribution with degrees of freedom = 38, is 2.025 which is more than the calculated value 1.94 of t. Hence, H0 accepted.[note: being a two-sided test, 5% point is used for 10% level of significance]
=> µ1 = µ2 => there is no evidence to suggest that CD writer brings down the battery life.
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