Suppose that in a random selection of 100 colored​ candies, 24​% of them are blue. the candy company claims that the percentage of blue candies is equal to 28​%. use a 0.10 significance level to test that claim.

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Answer:

Null hypothesis: p= 0.28

Alternative hypothesis: p not equal to 0.28

[tex] z=\frac{0.24 - 0.28}{\sqrt{0.28(1-0.28)/100}} [/tex]

[tex] z=-0.89 [/tex]

Critical value at [tex] \alpha =0.10 [/tex] is 1.645

Since, z=-0.89 lies between -1.645 and 1.645 we fail to reject the null hypothesis.

The [tex]z=-0.89[/tex] lies between [tex]-1.645[/tex] and [tex]1.645[/tex] we fail to reject the null hypothesis.

Step-by-step explanation:

Given: Suppose that in a random selection of [tex]100[/tex] colored​ candies, [tex]24\%[/tex] of them are blue. the candy company claims that the percentage of blue candies is equal to [tex]28\%[/tex].

According to question:

Null hypothesis: [tex]p=\frac{28}{100}=0.28[/tex]

Alternative hypothesis: not equal to

Now, [tex]z-\rm{value[/tex] is caluculated as:

[tex]z=\frac{0.24-0.28}{\sqrt{{0.28(1-0.28)}/{100}}}\\z=\frac{-0.04}{0.0448998}\\z=-0.89[/tex]

Critical value at  [tex]\alpha =0.10[/tex] is [tex]1.645[/tex]

Therefore, [tex]z=-0.89[/tex] lies between [tex]-1.645[/tex] and [tex]1.645[/tex] we fail to reject the null hypothesis.

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