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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

Respuesta :

Answer:  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 : [tex]H_0:p=P[/tex]

Alternate hypothesis : [tex]H_1:p\neq P[/tex]

So, the test statistic would be

[tex]z=\dfrac{p-P}{\sqrt{\dfrac{P.Q}{n}}}\\\\z=\dfrac{0.26-0.27}{\sqrt{\dfrac{0.27\times 0.73}{100}}}\\\\z=-0.2252[/tex]

Since α = 0.05

So, critical value of z = 1.96

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

Using the excel function , we get that

[tex]P(z>0.2252)\\\\=2\times 0.410.911845\\\\=0.8219[/tex]

Hence, a) -0.2252, b) 0.8219

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