Question

The owner of a pizza shop in Lincoln wants to know how many pizzas she will need to make this month. She has asked you to figure out how many she should be prepared to make for the upcoming month. She has been collecting data for the past year on how many pizzas she makes every month. The data is given below.

80, 67, 76, 56, 65, 90, 78, 77, 89, 83, 86, 74

83, 65, 77, 66, 75, 89, 87, 77, 89, 83, 83, 75

1. What is the confidence level used to make your confidence interval?

2. Interpret the confidence interval you created in 3).

3. The pizza shop owner believes she will not need to make more than 70 pizzas this month. Set up null and alternative hypotheses for this claim, and make a decision based on the evidence you have found with your confidence interval.

4. Change the alpha level to a=0.01 and repeat 2-4. How does this effect the confidence interval you created in 3)? Did you expect this? Why or why not?

5. What would you expect to happen to your confidence interval in 6) if you changed the sample standard deviation to 13? (you do not need to create a new confidence interval for this problem.)

Answer 1

1.
TRADITIONAL METHOD
given that,
sample mean, x =77.916
standard deviation, s =8.859
sample size, n =24
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 8.859/ sqrt ( 24) )
= 1.808
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 = 23 d.f is 2.069
margin of error = 2.069 * 1.808
= 3.741
III.
CI = x ± margin of error
confidence interval = [ 77.916 ± 3.741 ]
= [ 74.175 , 81.657 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =77.916
standard deviation, s =8.859
sample size, n =24
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 23 d.f is 2.069
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 = [ 77.916 ± t a/2 ( 8.859/ Sqrt ( 24) ]
= [ 77.916-(2.069 * 1.808) , 77.916+(2.069 * 1.808) ]
= [ 74.175 , 81.657 ]
-----------------------------------------------------------------------------------------------
2.
interpretations:
1) we are 95% sure that the interval [ 74.175 , 81.657 ] 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
3.
Given that,
population mean(u)=70
sample mean, x =77.916
standard deviation, s =8.859
number (n)=24
null, Ho: μ=70
alternate, H1: μ>70
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.714
since our test is right-tailed
reject Ho, if to > 1.714
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =77.916-70/(8.859/sqrt(24))
to =4.378
| to | =4.378
critical value
the value of |t α| with n-1 = 23 d.f is 1.714
we got |to| =4.378 & | t α | =1.714
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 4.3775 ) = 0.00011
hence value of p0.05 > 0.00011,here we reject Ho
ANSWERS
---------------
null, Ho: μ=70
alternate, H1: μ>70
test statistic: 4.378
critical value: 1.714
decision: reject Ho
p-value: 0.00011
we have enough evidence to support the claim that The pizza shop owner believes she will not need to make more than 70 pizzas this month.
4.
i.
Given that,
population mean(u)=70
sample mean, x =77.916
standard deviation, s =8.859
number (n)=24
null, Ho: μ=70
alternate, H1: μ>70
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.5
since our test is right-tailed
reject Ho, if to > 2.5
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =77.916-70/(8.859/sqrt(24))
to =4.378
| to | =4.378
critical value
the value of |t α| with n-1 = 23 d.f is 2.5
we got |to| =4.378 & | t α | =2.5
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 4.3775 ) = 0.00011
hence value of p0.01 > 0.00011,here we reject Ho
ANSWERS
---------------
null, Ho: μ=70
alternate, H1: μ>70
test statistic: 4.378
critical value: 2.5
decision: reject Ho
p-value: 0.00011
we have enough evidence to support the claim that The pizza shop owner believes she will not need to make more than 70 pizzas this month.
ii.
TRADITIONAL METHOD
given that,
sample mean, x =77.916
standard deviation, s =8.859
sample size, n =24
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 8.859/ sqrt ( 24) )
= 1.808
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 23 d.f is 2.807
margin of error = 2.807 * 1.808
= 5.076
III.
CI = x ± margin of error
confidence interval = [ 77.916 ± 5.076 ]
= [ 72.84 , 82.992 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =77.916
standard deviation, s =8.859
sample size, n =24
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 23 d.f is 2.807
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 = [ 77.916 ± t a/2 ( 8.859/ Sqrt ( 24) ]
= [ 77.916-(2.807 * 1.808) , 77.916+(2.807 * 1.808) ]
= [ 72.84 , 82.992 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 72.84 , 82.992 ] 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
5.
if sample standard deviation changes to 13 then,confidence interval
TRADITIONAL METHOD
given that,
sample mean, x =77.916
standard deviation, s =13
sample size, n =24
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 13/ sqrt ( 24) )
= 2.654
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 23 d.f is 2.807
margin of error = 2.807 * 2.654
= 7.449
III.
CI = x ± margin of error
confidence interval = [ 77.916 ± 7.449 ]
= [ 70.467 , 85.365 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =77.916
standard deviation, s =13
sample size, n =24
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 23 d.f is 2.807
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 = [ 77.916 ± t a/2 ( 13/ Sqrt ( 24) ]
= [ 77.916-(2.807 * 2.654) , 77.916+(2.807 * 2.654) ]
= [ 70.467 , 85.365 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 70.467 , 85.365 ] 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
The width of the confidence interval decreases as the sample size increases. The width increases as the standard deviation increases.