Suppose that in a random selection of 100 colored​ candies, 21​% of them are blue. The candy company claims that the percentage of blue candies is equal to 28​%. Use a 0.05 significance level to test that claim.

Respuesta :

Answer:

The percentage of blue candies is equal to 28​%.

Step-by-step explanation:

Sample size = n = 1000

21​% of them are blue

So, No. of blue candies = [tex]21\% \times 100 =\frac{21}{100} \times 100=21[/tex]

Claim : The percentage of blue candies is equal to 28​%.

[tex]H_0:\mu = 0.28\\H_a:\mu \neq 0.28[/tex]

We will use one sample proportion test  

[tex]\widehat{p}=\frac{x}{n}[/tex]

[tex]\widehat{p}=\frac{21}{100}[/tex]

[tex]\widehat{p}=0.21[/tex]

Formula of test statistic =[tex]\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

                                       =[tex]\frac{0.21-0.28}{\sqrt{\frac{0.28(1-0.28)}{100}}}[/tex]

                                       =−1.55

Now refer the p value from the z table

p value =0.0606

α =0.05

So, p value >  α

So, we failed to reject null hypothesis

So, the percentage of blue candies is equal to 28​%.