A candy company claims that its jelly bean mix contains 15% blue jelly beans. Suppose that the candies are packaged at random in small bags containing about 200 jelly beans. If you receive a bag with 40 blue jelly beans, would you be doubtful of the company's claim? Answer by calculating the appropriate z-score. Round to the nearest hundredth when necessary.

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Answer:

Doubtful.

Step-by-step explanation:

Our values are,

[tex]P=0.15 \rightarrow 15\%[/tex]Jelly Bean Mix.

The sample for us is n=200

And our proportion (x) is 40 .

We can calculate  with this dates the sample proportion and make he test statistic.

[tex]p=\frac{x}{n} = \frac{40}{200} = 0.20[/tex]

[tex]H_0:P=0.15[/tex]

[tex]H_1:P \neq 0.15[/tex]

We use the formula for Test statistic, that is given by,

[tex]z= \frac{p-P}{\sqrt{\frac{P-(1-P)}{n}}}[/tex]

[tex]z= \frac{0.20-0.15}{\sqrt{\frac{0.15(1-0.15)}{200}}}[/tex]

[tex]z= 1.98[/tex]

We see that our P-value: 0.048 (From the Table)

We note that p-value is less than 0.05, so we should to rejecyt the null hypothesis and conclude that the calim is doubtful