Pedro thinks that he has a special relationship with the number 5. In particular, Pedro thinks that he would roll a 5 with a fair 6-sided die more often than you'd expect by chance alone. Suppose p is the true proportion of the time Pedro will roll a 5. (Note you will need to find the probability of getting a four when rolling a fair 6-sided die like we did in chapter 3 so that you can compare p to this value.) (a) State the null and alternative hypotheses for testing Pedro's claim. (Type the symbol "p" for the population proportion, whichever symbols you need of "<", ">", "=", "not =", ">=", or "<=" and express any values as a fraction e. g. p = 1/3) H0 = Ha =
(b) Now suppose Pedro makes n = 38 rolls, and a 5 comes up 8 times out of the 38 rolls. Determine the P-value of the test: P-value =
(c) Answer the question: Does this sample provide evidence at the 5 percent level that Pedro rolls a 5 more often than you'd expect? (Type: Yes or No)

1.)

H0 : p = 1/6

H1 : p > 1/6

2.)

Pvalue = 0.3029

3.)

No

Step-by-step explanation:

Probability of rolling 5 in a 6-sided die by chance :

Number of 5's = 1

Number of 6 = 6

p = 1/6

The Null hypothesis :

H0 : p = 1/6

Since Pedro thinks he'd roll more than just by chance ;

Alternative hypothesis :

H1 : p > 1/6

Given that :

n = 38 ` x = 5

Phat = x / n = 5 / 38

Teat statistic : (phat - p) / √p(1-p)/n

1-p = 1 - 1/6 = 5/6

phat - p = 5/38 - 1/6 = −0.035087

√p(1-p) /n = (1/6 * 5/6) / 30 = √0.0046296 = 0.0680

Test statistic = −0.035087 / √0.0046296 =

Test statistic = −0.515673

Using the Pvalue calculator :

Pvalue at 0.05 , 1 - tail = 0.3029

Pvalue = 0.3029

Hence, at α = 0.05

Pvalue > α ;

There is no evidence to support Pedro's claim

answer: 2 answers

step-by-step explanation: the first one means to spread the product throughout the marketplace such that a large number of people can buy it.  

the second means the process of making a product or service available for the consumer or business user that needs it.

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