A candy company makes five different colors of candies. Find the sample proporti

Question

A candy company makes five different colors of candies. Find the sample proportion of candies that are green. Use that result to test the claim that 17% of the company's candies are green. Use the data in the table below, the P-value method, and a significance level of 0.01.

A) What is the test statistic

B) Determine the p-value

C) Make final conclusion

Red Blue Brown Green Yellow 0.821 0.941 0.924 0.955 0.999 0.995 0.725 0.884 0.768 0.728 0.955 0.888 0.859 0.933 0.743 0.785 0.865 0.898 0.766 0.879 0.922 0.899 0.998 0.705 0.883 0.975 0.934 0.777 0.701 0.732 0.721 0.714 0.716 0.902 0.863 0.858 0.984 0.728 0.744 0.825 0.833 0.919 0.842 0.796 0.805 0.878 0.884 0.856 0.878 0.934 0.737 0.876 0.831 0.906 0.931 0.851 0.815 0.802 0.787 0.766 0.776 0.751 0.928 0.988 0.745 0.828 0.736 0.912 0.878 0.872 0.942 0.777 0.949 0.912 0.847 0.923 0.925 0.848 0.933 0.956 0.864

Explanation / Answer

Answer to the question)

We got 15 cadies green out of toal 81 candies

Sample proporiton p^ = 15/81 = 18.52% or 0.1852

.

Population Proporiton P = 0.17

.

(it is a two tailed test)

.

SE = sqrt(0.17*0.83/81) = 0.0417

.

Formula of test statistic

z = (p^ - P) / SE

z = (0.1852-0.15) / 0.0417

z = 0.844125

.

P value = 0.3986 .[ I used excel command =2*(1-NORMSDIST(0.844125)) to find the P value ]

.

Inference: Since the P value 0.3986 > alpha 0.05 , we fail to reject the null hypothesis

.

Conclusion: Thus we conclude that the claim that the green candies proportion is 17% is correct

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