Question

A machine in the student lounge dispenses coffee. The average cup of coffee is supposed to contain 7.0 ounces. A random sample of twelve cups of coffee from this machine show the average content to be 7.2 ounces with a standard deviation of 0.60 ounce. Do you think that the machine has slipped out of adjustment and that the average amount of coffee per cup is different from 7 ounces? Use a 5% level of significance.

What are we testing in this problem?

single mean

single proportion    

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: p = 7; H₁: p < 7

H₀: p = 7; H₁: p ≠ 7   

H₀: p = 7; H₁: p > 7

H₀: μ = 7; H₁: μ > 7

H₀: μ = 7; H₁: μ < 7

H₀: μ = 7; H₁: μ ≠ 7

(b) What sampling distribution will you use? What assumptions are you making?

The standard normal, since we assume that x has a normal distribution with unknown σ.

The standard normal, since we assume that x has a normal distribution with known σ.    

The Student's t, since we assume that x has a normal distribution with unknown σ.

The Student's t, since n is large with unknown σ.

What is the value of the sample test statistic? (Round your answer to three decimal places.)


(c) Find (or estimate) the P-value.

P-value > 0.50

00.250 < P-value < 0.500   

0.100 < P-value < 0.250

0.050 < P-value < 0.100

0.010 < P-value < 0.050

P-value < 0.010

Sketch the sampling distribution and show the area corresponding to the P-value.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.   

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

(e) Interpret your conclusion in the context of the application.

There is sufficient evidence at the 0.05 level to conclude that the mean amount of coffee per cup differs from 7 ounces.

There is insufficient evidence at the 0.05 level to conclude that the mean amount of coffee per cup differs from 7 ounces.    

Answer 1

Given that,
population mean(u)=7
sample mean, x =7.2
standard deviation, s =0.6
number (n)=12
null, Ho: μ=7
alternate, H1: μ!=7
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.201
since our test is two-tailed
reject Ho, if to < -2.201 OR if to > 2.201
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =7.2-7/(0.6/sqrt(12))
to =1.1547
| to | =1.1547
critical value
the value of |t α| with n-1 = 11 d.f is 2.201
we got |to| =1.1547 & | t α | =2.201
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.1547 ) = 0.2727
hence value of p0.05 < 0.2727,here we do not reject Ho
ANSWERS
---------------
t test for single mean with unknown population standard deviation
a.
null, Ho: μ=7
alternate, H1: μ!=7
b.
The standard normal, since we assume that x has a normal distribution with unknown σ.
c.
test statistic: 1.1547
critical value: -2.201 , 2.201
decision: do not reject Ho
p-value: 0.2727
00.250 < P-value < 0.500
d.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
e.
we do not have enough evidence to support the claim that the machine has slipped out of adjustment and that the average amount of coffee per cup is different from 7 ounces