You are hired by a cell phone provider to determine the number of times per day

Question

You are hired by a cell phone provider to determine the number of times per day their customers interact with their smartphone. You take a random sample of 28 customers and find the average number of interactions to be 60 times per day with a sample standard deviation of 6. Show all work. Be sure to include your probability and calculator statements.

a. If you are to find a confidence interval for the population mean number of interactions, what distribution would you use and why?

b. Find a 99% confidence interval estimate for the population mean number of interactions.

c. What is the margin of error for the confidence interval you found in part b?

d. A representative for the cell phone provider told you they believe the mean number of interactions per day was 65. What would you say to the company representative? Explain.

Explanation / Answer

a.
we are using normal distribution with the data,
we are using 99% confidence interval for t test single mean because they given in that data sample stanadard deviation
b.
TRADITIONAL METHOD
given that,
sample mean, x =60
standard deviation, s =6
sample size, n =28
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 6/ sqrt ( 28) )
= 1.134
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 27 d.f is 2.771
margin of error = 2.771 * 1.134
= 3.142
III.
CI = x ± margin of error
confidence interval = [ 60 ± 3.142 ]
= [ 56.858 , 63.142 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =60
standard deviation, s =6
sample size, n =28
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 27 d.f is 2.771
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 = [ 60 ± t a/2 ( 6/ Sqrt ( 28) ]
= [ 60-(2.771 * 1.134) , 60+(2.771 * 1.134) ]
= [ 56.858 , 63.142 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 56.858 , 63.142 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean

c.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 27 d.f is 2.771
margin of error = 2.771 * 1.134
= 3.142

d.
A representative for the cell phone provider told you they believe the mean number of interactions per day was 65.
Given that,
population mean(u)=65
sample mean, x =60
standard deviation, s =6
number (n)=28
null, Ho: =65
alternate, H1: !=65
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.05
since our test is two-tailed
reject Ho, if to < -2.05 OR if to > 2.05
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =60-65/(6/sqrt(28))
to =-4.41
| to | =4.41
critical value
the value of |t | with n-1 = 27 d.f is 2.05
we got |to| =4.41 & | t | =2.05
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -4.4096 ) = 0.0001
hence value of p0.05 > 0.0001,here we reject Ho
ANSWERS
---------------
null, Ho: =65
alternate, H1: !=65
test statistic: -4.41
critical value: -2.05 , 2.05
decision: reject Ho
p-value: 0.0001
we have enough evidence to support the claim that A representative for the cell phone provider told you they believe the mean number of interactions per day was 65

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