Chapter 11.2, Problem 32E

a)

To determine

To verify the conditions for inference.

Answer to Problem 32E

All conditions satisfied.

Explanation of Solution

Given:

		Water temperature
		Cold	Neutral	Hot	Total
Hatching status	Yes	16	38	75	129
	No	11	18	29	58
	Total	27	56	104	187

Calculation:

The conditions of chi square test are: Random, Independence and Large counts.

The eggs randomly assigned to a water temperature, so condition of random sample is satisfied.

The sample size is 187 which is less than 10% of all eggs so, the condition of independent is satisfied.

The expected count of each survey is at least 5 so, condition of large count is satisfied.

b)

To determine

To find p-value using table

Answer to Problem 32E

P-value > 0.25

Explanation of Solution

Given:

		Water temperature
		Cold	Neutral	Hot	Total
Hatching status	Yes	16	38	75	129
	No	11	18	29	58
	Total	27	56	104	187

The value of chi square test statistic is 1.7033

Formula:

The degrees of freedom = $(r-1)(c-1)$

Calculation:

The degrees of freedom = $(2-1)(3-1)=2$

Using table, p-value is

p > 0.25

c)

To determine

To find p-value using calculator

Answer to Problem 32E

P-value is 0.4267

Explanation of Solution

Given:

		Water temperature
		Cold	Neutral	Hot	Total
Hatching status	Yes	16	38	75	129
	No	11	18	29	58
	Total	27	56	104	187

The value of chi square test statistic is 1.7033

Formula:

The degrees of freedom = $(r-1)(c-1)$

Calculation:

The degrees of freedom = $(2-1)(3-1)=2$

Using excel formula, =CHIDIST(1.7033,2)

P-value = 0.4267 = 42.67%

Therefore, there is 42.67% chance of obtaining the sample results of more extreme, when the distribution of hatching are not same for each temperature.

d)

To determine

To explain conclusion.

Answer to Problem 32E

There is no convincing evidence that the distribution of hatching is not the same for each water temperature.

Explanation of Solution

Given:

		Water temperature
		Cold	Neutral	Hot	Total
Hatching status	Yes	16	38	75	129
	No	11	18	29	58
	Total	27	56	104	187

The value of chi square test statistic is 1.7033

P-value = 0.4267

Calculation:

Decision: P-value > 0.10, fail to reject H0.

Conclusion: There is no convincing evidence that the distribution of hatching is not the same for each water temperature.