Chloe breaks into the pantry after Albert goes to work. She finds and opens a family-pack of mixed nuts. After surfing the web, she finds that these packs claim to be 20% almond (A), 30% cashew (C), 10% macadamia (M), and 40% peanut. Chloe discovers that her bag contains 14 almonds, 28 cashews, 6 macadamias, and 52 peanuts. a) Perform an appropriate test to determine whether this claim is true or not at a = 0.05. State all the relevant steps (listed in exercise 1).
b) What is the decision and conclusion (write a sentence) at a = 0.1?
c) Chloe is a choosy and sneaky doggie. She likes to find a p-value first and then come up with a significance level, a later so she can reject as many Hos as possible. Is her method acceptable?

Solution :

In this case we have to use for the goodness of fit.

Null hypothesis is : Data follows the given distribution

Alternate hypothesis is : Data do not follow the given distribution

The level of significance,

The test statistics is given by :

Chi square =

Here O is the observed frequencies

         E is the expected frequencies

So we have, N = number of the categories = 4

df = degrees o freedom = N - 1

                                        = 4 - 1 = 3

Critical value  =

Calculating the table for the test statistics are given as :

  Category      prop.     O        E      

   A                   0.2     14       20         1.80

   C                   0.3     28       30         0.13

   M                  0.1        6        10          1.60

  P                    0.4      52       40         3.60

Total                  1        100      100        7.13

The test statics = chi square  = = 7.13

statistics = 7.13

P-value = 0.067767248

P -value >

Therefore, we do not reject the . There is sufficient evidence to conclude that the data follows the given distribution. So there is sufficient evidence to conclude the given claim is true.

68719476736

step-by-step explanation:

8*8*8*8*8*8*8*8*8*8*8*8=68719476736