Question

Problem 1.

1. A local coffee shop is concerned whether their customers are getting what they pay for. For a “medium” sized coffee, you’re supposed to get 10 ounces, but the amount varies slightly from cup to cup. A random sample of 20 cups was taken, which has a mean 9.85 and a standard deviation .19. Is there evidence that the coffee shop is shortchanging customers?

a. What are the Parameter of interest and Hypotheses?

b. Are conditions met for inference procedures?

c. What are the degrees of freedom, df ?

d. What is the mean and standard deviation of the sample? With proper notation.

e. What is the standard error of the sample mean, se(y)?

f. Find the test statistic, t.

g. What is the p-value?

h. What is your conclusion, in context of the problem?

i. If instead the coffee shop was worried the machine was dispensing too much coffee, what would the null and alternative hypotheses be? What would the p-value be?

j. If the coffee shop was just interested in determine if there was a difference in the amount advertised and the amount the customers were receiving, what would the null and alternative hypotheses be? What would the p-value be?

2. Create a 95% confidence interval for the population parameter. We have already checked conditions for this situation.

a. What is the critical value, t*, for this interval?

b. What is the confidence interval? Interpret the interval in context of the problem

Please solve all parts thank you.

Answer 1

Given that,
population mean(u)=10
sample mean, x =9.85
standard deviation, s =0.19
number (n)=20
null, Ho: μ=10
alternate, H1: μ!=10
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =9.85-10/(0.19/sqrt(20))
to =-3.5306
| to | =3.5306
critical value
the value of |t α| with n-1 = 19 d.f is 2.093
we got |to| =3.5306 & | t α | =2.093
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.5306 ) = 0.0022
hence value of p0.05 > 0.0022,here we reject Ho
ANSWERS
---------------
a.
the Parameter of interest and Hypotheses is T test for single mean with unknown population standard deviation
b.
yes,
conditions met for inference procedures
c.
d.f is 2.093
d.
the mean = 9.85 and standard deviation of the sample =0.19
e.
the standard error of the sample mean
standard error = ( 0.19/ sqrt ( 20) )
= 0.042
null, Ho: μ=10
alternate, H1: μ!=10
f.
test statistic: -3.5306
critical value: -2.093 , 2.093
decision: reject Ho
g.
p-value: 0.0022
h.
we have enough evidence to support the claim that t the coffee shop is shortchanging customers.
i.
If instead the coffee shop was worried the machine was dispensing too much coffee, the null and alternative hypotheses be
p value is greater than alpha value
j.
if there was a difference in the amount advertised and the amount the customers were receiving, would the null and alternative hypotheses be
null, Ho: μ=10
alternate, H1: μ!=10
p-value: 0.0022
2.
TRADITIONAL METHOD
given that,
sample mean, x =9.85
standard deviation, s =0.19
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.19/ sqrt ( 20) )
= 0.042
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 = 19 d.f is 2.093
margin of error = 2.093 * 0.042
= 0.089
III.
CI = x ± margin of error
confidence interval = [ 9.85 ± 0.089 ]
= [ 9.761 , 9.939 ]
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DIRECT METHOD
given that,
sample mean, x =9.85
standard deviation, s =0.19
sample size, n =20
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.093
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 = [ 9.85 ± t a/2 ( 0.19/ Sqrt ( 20) ]
= [ 9.85-(2.093 * 0.042) , 9.85+(2.093 * 0.042) ]
= [ 9.761 , 9.939 ]
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interpretations:
1) we are 95% sure that the interval [ 9.761 , 9.939 ] 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
a.
critical value: -2.093 , 2.093
b.
95% sure that the interval [ 9.761 , 9.939 ]