Question

A wine shop wants to serve its white wine at a mean temperature of 50 degrees. In a sample of 36 glasses of wine, the mean was found to be 53 with a standard deviation of 3. The wine shop is interested to see if the evidence indicates that the wine shop is not meeting its temperature standard.

(a) Using a significance level of α=0.05, perform a two-tailed hypothesis test to determine if the wine shop is not meeting its temperature standard. Use the test statistic/critical value approach.

(b) Using a significance level of α=0.01, perform a two-tailed hypothesis test to determine if the wine shop is not meeting its temperature standard. Use the test statistic/critical value approach.

(c) Repeat part (a), but use the p-value approach.

(d) Repeat part (b), but use the p-value approach.

Answer 1

a.
Given that,
population mean(u)=50
sample mean, x =53
standard deviation, s =3
number (n)=36
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.03
since our test is two-tailed
reject Ho, if to < -2.03 OR if to > 2.03
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =53-50/(3/sqrt(36))
to =6
| to | =6
critical value
the value of |t α| with n-1 = 35 d.f is 2.03
we got |to| =6 & | t α | =2.03
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: 6
critical value: -2.03 , 2.03

b.
Given that,
population mean(u)=50
sample mean, x =53
standard deviation, s =3
number (n)=36
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =2.72
since our test is two-tailed
reject Ho, if to < -2.72 OR if to > 2.72
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =53-50/(3/sqrt(36))
to =6
| to | =6
critical value
the value of |t α| with n-1 = 35 d.f is 2.72
we got |to| =6 & | t α | =2.72
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: 6
critical value: -2.72 , 2.72

c.
Given that,
population mean(u)=50
sample mean, x =53
standard deviation, s =3
number (n)=36
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.03
since our test is two-tailed
reject Ho, if to < -2.03 OR if to > 2.03
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =53-50/(3/sqrt(36))
to =6
| to | =6
critical value
the value of |t α| with n-1 = 35 d.f is 2.03
we got |to| =6 & | t α | =2.03
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 6 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: 6
critical value: -2.03 , 2.03
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that the wine shop is not meeting its temperature standard.

d.
Given that,
population mean(u)=50
sample mean, x =53
standard deviation, s =3
number (n)=36
null, Ho: μ=50
alternate, H1: μ!=50
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =2.72
since our test is two-tailed
reject Ho, if to < -2.72 OR if to > 2.72
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =53-50/(3/sqrt(36))
to =6
| to | =6
critical value
the value of |t α| with n-1 = 35 d.f is 2.72
we got |to| =6 & | t α | =2.72
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 6 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=50
alternate, H1: μ!=50
test statistic: 6
critical value: -2.72 , 2.72
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that the wine shop is not meeting its temperature standard.