Question

A magazine collects data each year on the price of a hamburger in a certain fast food restaurant in various countries around the world. The price of this hamburger for a sample of restaurants in Europe in January resulted in the following hamburger prices (after conversion to U.S. dollars). 5.18 4.99 4.06 4.68 5.25 4.66 4.12 4.98 5.19 5.59 5.33 4.60 The mean price of this hamburger in the U.S. in January was $4.61. For purposes of this exercise, assume it is reasonable to regard the sample as representative of these European restaurants. Does the sample provide convincing evidence that the mean January price of this hamburger in Europe is greater than the reported U.S. price? Test the relevant hypotheses using α = 0.05. (Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places.)

t =

P-value =

Answer 1

We have to test "Does the sample provide convincing evidence that the mean January price of this hamburger in Europe is greater than the reported U.S. price?"

hypothesis :

( claim )

Right tailed test.

Test statistics -

Where, is sample mean.

is hypothesized mean.

s is sample standard deviation.

n is sample size.

From given data,

= 4.885833 , s = 0.473679 , n = 12 , = 4.61

So test statistics is,

t =2.01722

P-value -

This is right tailed test. So p-value for this test is given by,

P-value = P( t > 2.01722 )

Using Excel function,    =TDIST( t, df, tails )      df = n - 1 = 12 - 1 = 11

p-value = TDIST( 2.01722 , 11 , 1 ) = 0.034372

P-value = 0.0344

Decision about null hypothesis -

We are given , = 0.05

P-value is less than = 0.05

So reject null hypothesis.

Conclusion -

There is sufficient evidence to say that the mean January price of this hamburger in Europe is greater than the reported U.S. price