Question

degrees of freedom for either the numerator or denominator of F, use the closest smaller number. For instance, if the degrees of freedom for the denominator were df, 23 you would use the critical value for df = 20 (2,3) 4. Chapter 10, Problem 10.10 If the degrees of freedom for I do not appear in Table A.2 use the closest smaller degrees of freedom to determine the critical value of t. For instance, if the degrees of freedom are 50 use the critical value of t for degrees of freedom equal to 40. (4,4) 5. Chapter 10, Problem 10.13 In addition to the requested test, construct a 95 percent confidence interval on the
please help with question 4

Answer 1

Let $\mu_L$ and $\mu_S$ be the mean standard length of largemouth bass and smallmouth bass respectively.

Null hypothesis H0:

L = us

Alternative hypothesis H0:

Srl + In

Since we do not know the population standard deviations and do not assume that the population standard deviations of the two groups are equal, we will use independent samples t test.

The  standard error (SE) of the sampling distribution of difference in means is,

SE = sqrt[ (s₁²/n₁) + (s₂²/n₂) ] = sqrt[ (96.4²/125) + (40²/97) ] = 9.53

The degrees of freedom (DF) is:

DF = (s₁²/n₁ + s₂²/n₂)² / { [ (s₁² / n₁)² / (n₁ - 1) ] + [ (s₂² / n₂)² / (n₂ - 1) ] }

= (96.4²/125 + 40²/97)² / { [ (96.4² / 125)² / (125 - 1) ] + [ (40² / 97)² / (97 - 1) ] }

= 174 (Rounding to nearest integer)

For $\alpha$ = 0.05, the critical value of t at DF = 174 is 1.984 (We take the value at DF = 100)

Test statistic, t = ($\bar{x_1}$ - $\bar{x_2}$ ) / SE = (272.8 - 164.8) / 9.53 = 11.33

Since the test statistic is greater than the critical value, we reject null hypothesis H0 and conclude that there is significant evidence that or mean standard length of largemouth bass and smallmouth bass differ significantly.

Srl + In

Margin of error = t * SE = 1.984 * 9.53 = 18.91

Sample mean difference = $\bar{x_1}$ - $\bar{x_2}$ = 272.8 - 164.8 = 108

95% confidence interval of mean difference is,

(108 - 18.91, 108 + 18.91)

= (89.09,  126.91)