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 = =p
The formula for the standard deviation of a proportion is
Solving for the standard deviation gives;
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× = 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;
Therefore, the z-score more than 2 × which is significant.