1) Your company is evaluating two cloud-based secured data storage services. “Pi
ID: 3314639 • Letter: 1
Question
1)
Your company is evaluating two cloud-based secured data storage services. “Pie in the Sky,” the newer service, claims its uploading and downloading speeds are faster than the older service, “Cloudy but Steady Skies.” You need to make a decision based on published access times for both services at different times and for varying file sizes. To make your decision, you purchase a statistical study, which indicates that average download time for Pie in the Sky is 0.77 sec. per MB and for Cloudy but Steady Skies is 0.84. Assume that n1 = n2 = 50,1 = 0.2 and 2 = 0.3. With = .05, the appropriate decision is ______.
reject the null hypothesis 12 < 22
accept the alternate hypothesis 1 2 > 0
reject the alternate hypothesis n1 = n2 = 50
fail to reject the null hypothesis 1 2 = 0
do nothing
2)
A researcher is estimating the average difference between two population means based on matched pairs samples. She gathers data on each pair in the study resulting in:
Assume that the data are normally distributed in the population. A 95% confidence interval would be _______.
3)
You are evaluating investing in a cognitive training company. For this reason, you want to determine whether users who complete at least 75% of the recommended daily training for two months show improved levels of reading comprehension and problem solving skills. You select a random sample of new users and get test scores for each participant in the sample. The test score is a composite of reading comprehension and problem solving skills. Two months later, you randomly select 15 users who have completed 75% or more of the recommended training for the last two months and have them take a test similar to the initial test. You are interested in determining whether the average test score before training isdifferent than the average test score after training for this sample. The after-training average is 92.8, which is 2.7 points higher than the before-training average. The sample standard deviations of the differences is 1.2. You use a significance level of 0.10, and you can assume that the differences are normally distributed in the population. The table t-value for this test is ______.
1.761
1.746
1.753
1.345
1.339
reject the null hypothesis 12 < 22
accept the alternate hypothesis 1 2 > 0
reject the alternate hypothesis n1 = n2 = 50
fail to reject the null hypothesis 1 2 = 0
do nothing
2)
A researcher is estimating the average difference between two population means based on matched pairs samples. She gathers data on each pair in the study resulting in:
Assume that the data are normally distributed in the population. A 95% confidence interval would be _______.
3.02 to 0.18
1.6 to 1.09
2.11 to 1.09
2.11 to 1.09
3.23 to 2.23
3)
You are evaluating investing in a cognitive training company. For this reason, you want to determine whether users who complete at least 75% of the recommended daily training for two months show improved levels of reading comprehension and problem solving skills. You select a random sample of new users and get test scores for each participant in the sample. The test score is a composite of reading comprehension and problem solving skills. Two months later, you randomly select 15 users who have completed 75% or more of the recommended training for the last two months and have them take a test similar to the initial test. You are interested in determining whether the average test score before training isdifferent than the average test score after training for this sample. The after-training average is 92.8, which is 2.7 points higher than the before-training average. The sample standard deviations of the differences is 1.2. You use a significance level of 0.10, and you can assume that the differences are normally distributed in the population. The table t-value for this test is ______.
1.761
1.746
1.753
1.345
1.339
Explanation / Answer
Q1.
Given that,
mean(x)=0.77
standard deviation , 1 =0.2
number(n1)=50
y(mean)=0.84
standard deviation, 2 =0.3
number(n2)=50
null, Ho: u1 - u2 = 0
alternate, H1: 1 - u2 < 0
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=0.77-0.84/sqrt((0.04/50)+(0.09/50))
zo =-1.37
| zo | =1.37
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =1.373 & | z | =1.645
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -1.37 ) = 0.08491
hence value of p0.05 < 0.08491,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 - u2 = 0
alternate, 1 - u2 < 0
test statistic: -1.37
critical value: -1.645
decision: do not reject Ho
p-value: 0.08491
fail to reject the null hypothesis 1 2 = 0
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