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The Acme Candy Company claims that​ 60% of the jawbreakers it produces weigh more than 0.4 ounces. Suppose that 800 jawbreakers are selected at random from the production lines. Would it be significant for this sample of 800 to contain 494 jawbreakers that weigh more than 0.4​ ounces? Consider as significant any result that differs from the mean by at least 2 standard deviations. That​ is, significant values are either less than or equal to muminus2sigma or greater than or equal to muplus2sigma.

Respuesta :

Answer:

Yes, it would be statistically significant

Step-by-step explanation:

The information given are;

The percentage of jawbreakers it produces that weigh more than 0.4 ounces = 60%

Number of jawbreakers in the sample, n = 800

The mean proportion of jawbreakers that weigh more than 0.4 = 60% = 0.6 = [tex]\mu _ {\hat p}[/tex] =p

The formula for the standard deviation of a proportion is [tex]\sigma _{\hat p} =\sqrt{\dfrac{p(1-p)}{n} }[/tex]

Solving for the standard deviation gives;

[tex]\sigma _{\hat p} =\sqrt{\dfrac{0.6 \cdot (1-0.6)}{800} } = 0.0173[/tex]

Given that the mean proportion is 0.6, the expected value of jawbreakers that weigh more than 0.4  in the sample of 800 = 800*0.6 = 480

For statistical significance the difference from the mean = 2×[tex]\sigma _{\hat p}[/tex] = 2*0.0173 = 0.0346 the equivalent number of Jaw breakers = 800*0.0346 = 27.7

The z-score of 494 jawbreakers is given as follows;

[tex]Z=\dfrac{x-\mu _{\hat p} }{\sigma _{\hat p} }[/tex]

[tex]Z=\dfrac{494-480 }{0.0173 } = 230.94[/tex]

Therefore, the z-score more than 2 ×[tex]\sigma _{\hat p}[/tex] which is significant.

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