Question

Packages of mixed nuts made by a certain company contain four types of nuts. The percentages of nuts of Types 1, 2, 3, and 4 are advertised to be 40%, 30%, 20%, and 10%, respectively. A random sample of nuts is selected, and each one is categorized by type.

(a)

If the sample size is

n = 200

and the resulting test statistic value is

X² = 17.5,

what conclusion would be appropriate for a significance level of 0.001? (Use a table or technology.)

The P-value is less than the significance level of α = 0.001, so we reject H₀. We have convincing evidence that the percentages of nuts of the different types are not as advertised.The P-value is greater than the significance level of α = 0.001, so we fail to reject H₀. We have convincing evidence that the percentages of nuts of the different types are not as advertised.    The P-value is greater than the significance level of α = 0.001, so we fail to reject H₀. We do not have convincing evidence that the percentages of nuts of the different types are not as advertised.The P-value is less than the significance level of α = 0.001, so we reject H₀. We do not have convincing evidence that the percentages of nuts of the different types are not as advertised.

(b)

If the random sample had consisted of only 40 nuts, would you use the chi-square goodness-of-fit test? Explain your reasoning.

If the sample consisted of only 40 nuts,

the expected value for Type 1 =,

the expected value for Type 2=

the expected value for Type 3 =

the expected value for Type 4

Because the expected value for  ---Select--- every nut type at least one nut type  is  ---Select--- greater less  than 5, the sample size  ---Select--- is is not  large enough to use the chi-square goodness-of-fit test.

Answer 1

Solution :

a) The value of the test statistic is X² = 17.5.

Degrees of freedom = (k - 1) = (4 - 1) = 3

(Where, k is number of nut types.)

The p-value for the test statistic is given as follows :

P-value = P(X² > value of the test statistic)

P-value = P(X² > 17.5)

P-value = 0.0006

Significance level = 0.001

(0.0006 < 0.001)

Conclusion :

The P-value is less than the significance level of α = 0.001, so we reject H₀. We have convincing evidence that the percentages of nuts of the different types are not as advertised.

b) Sample size for nuts = 40

The expected value for Type 1 = (40 × 0.4) = 16

the expected value for Type 2 = (40 × 0.3) = 12

the expected value for Type 3 = (40 × 0.2) = 8

the expected value for Type 4 = (40 × 0.1) = 4

Because the expected value for at least one nut type  is less than 5, therefore sample size is not large enough to use the chi-square goodness-of-fit test.

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