Question

The technology committee at a college has stated that the average time spent by students per lab visit has​ increased, and the increase supports the need for increased lab fees. To substantiate this​ claim, the committee randomly samples 12 student lab visits and notes the amount of time spent using the computer. The times are given in the accompanying table.

Time(min) 55,57,54,55,60,66,57,50,141,78,5,65

The previous mean amount of time spent using the lab computer was 55 minutes. Find a​ 95% confidence interval for the true mean. Make a conclusion about the claim. If there are​ outliers, find intervals with and without the outliers present.

The​ 95% confidence interval including any outliers that may exist is ​(________, ________​) minutes.

​(Round to one decimal place as​ needed.)

Make a conclusion about the claim.

A. Because there are no outliers, the conditions for inference are met, and the confidence is entirely above 55 minutes, conclude that the mean is above 55 minutes.

B. Because there are no outliers, the conditions for inference are met, and the confidence interval includes 55 minutes, conclude that the new mean is above 55 minutes.

C. In either case, one would be reluctant to conclude that the new mean is above 55 minutes. the sample size is large and the presence of two large outliers is evidence that the assumptions for interference are satisfied.

D. In either case, one would be reluctant to conclude that the new mean is above 55 minutes. The ample size is small and the presence of two large outliers advises one to be cautious about conclusions from this sample.

Answer 1

i.
outliers are including
Given that,
population mean(u)=55
sample mean, x =61.916
standard deviation, s =29.055
number (n)=12
null, Ho: μ=55
alternate, H1: μ>55
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.796
since our test is right-tailed
reject Ho, if to > 1.796
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =61.916-55/(29.055/sqrt(12))
to =0.825
| to | =0.825
critical value
the value of |t α| with n-1 = 11 d.f is 1.796
we got |to| =0.825 & | t α | =1.796
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 0.8246 ) = 0.21357
hence value of p0.05 < 0.21357,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=55
alternate, H1: μ>55
test statistic: 0.825
critical value: 1.796
decision: do not reject Ho
p-value: 0.21357
we do not have enough evidence to support the claim that the mean is above 55 minutes.
option:D
D. In either case, one would be reluctant to conclude that the new mean is above 55 minutes.
The ample size is small and the presence of two large outliers advises one to be cautious about conclusions from this sample.
ii.
TRADITIONAL METHOD
given that,
sample mean, x =61.916
standard deviation, s =29.055
sample size, n =12
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 29.055/ sqrt ( 12) )
= 8.4
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 2.201
margin of error = 2.201 * 8.4
= 18.5
III.
CI = x ± margin of error
confidence interval = [ 61.916 ± 18.5 ]
= [ 43.5 , 80.4 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =61.916
standard deviation, s =29.055
sample size, n =12
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 2.201
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 61.916 ± t a/2 ( 29.055/ Sqrt ( 12) ]
= [ 61.916-(2.201 * 8.4) , 61.916+(2.201 * 8.4) ]
= [ 43.5 , 80.4 ]
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interpretations:
1) we are 95% sure that the interval [ 43.5 , 80.4 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean