Suppose that in a random selection of 100 colored candies, 26% of them are blue. the candy company claims that the percentage of blue candies is equal to 27%. use a 0 .05 significance level to test that claim. identify the test statistic for this hypothesis test. find the p-value for this hypothesis test

 a) -0.2252, b) 0.8219

Step-by-step explanation:

Since we have given that

Sample size n = 100

Probability that candies are blue = p= 0.26

Probability that company claims that it is blue candy = P = 0.27

So, Q = 1-P= 1-0.27 = 0.73

So, Null hypothesis :

Alternate hypothesis :

So, the test statistic would be

Since α = 0.05

So, critical value of z = 1.96

p-value = P(Z>Z(calculated)

Using the excel function , we get that

Hence, a) -0.2252, b) 0.8219

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.

understand that i don't know the answers

will someone answer his question