Question

Note. For each of the following problems, state the appropriate null hypothesis and the alternative hypoth- esis, conduct the appropriate test, and state a conclusion. Reject the Ho if p < 0.05. Be sure to include the p-value. Assume the variable in question to be approximately normally distributed unless instructed otherwise. Part I: Solve the following problems by hand 1. Random samples of largemouth bass (Micropterus salmoides) and smallmouth bass (M. dolomieu) were taken from a lake and their lengths (in millimeters) were determined. We wish to know if the mean standard length differs between the two species in this lake. The results were as follows (data from J. Kagel). Largemouth Bass Smallmouth Bass 272.8 164.8 40.0 125 97 96.4 2. Seven tomato plants (Solanum lycopersicum) were treated with chlorogenic acid to determine if this would influence the activity of the enzyme o-diphenol oxidase in the leaves. Enzyme activity was measured as ul O2/mg protein/minute. A control group of seven plants were not treated. We do not know if this variable is approximately normally distributed, nor is it possible to determine this with a sample of this small a size. Does this treatment affect activity of the enzyme? Treated Untreated

Answer 1

1

Null hypothesis - There is no significant difference in mean length of the small mouth bass and large mouth bass

Alternate hypothesis - There is a significant difference in mean length of the small mouth bass and large mouth bass

For large mouth bass -

mean = 272.8

standard deviation = 96.4

sample size (n1) = 125

standard error (se1) = standard deviation / sqrt (sample size)

= 8.17

For large mouth bass -

mean = 164.8

standard deviation = 40

sample size (n2) = 97

standard error (se2) = standard deviation / sqrt (sample size)

= 4.06

Significance level (alpha) = 0.05

degrees of freedom (df) = n1 + n2 - 2

= 220

standard error of the difference = sqrt [ (se1^2 + se2^2) / 2]

= 9.12

test statistic = mean1 - mean2 / standard error

= 11.84

P value for t statistic of 11.84 and df 220 is less than 0.00001 (from normal distribution table)

Since the p value is less than 0.05, we conclude that

There is a significant difference in mean length of the small mouth bass and large mouth bass

2

Null hypothesis - There is no significant difference in mean enzyme activity between control and treatment group

Alternate hypothesis - There is a significant difference in mean enzyme activity between control and treatment group

For control group -

mean = 32.14

standard deviation = 10

sample size (n1) = 7

standard error (se1) = standard deviation / sqrt (sample size)

= 3.77

For treatment group -

mean = 14.4

standard deviation = 7.6

sample size (n2) = 7

standard error (se2) = standard deviation / sqrt (sample size)

= 5.43

Significance level (alpha) = 0.05

degrees of freedom (df) = n1 + n2 - 2

= 12

standard error of the difference = sqrt [ (se1^2 + se2^2) / 2]

= 6.61

test statistic = mean1 - mean2 / standard error

= 2.68

P value for t statistic of 2.68 and df 12 is 0.02 (from normal distribution table)

Since the p value is less than 0.05, we conclude that

There is a significant difference in mean enzyme activity between control and treatment group

Or, this treatment affects the activity of the enzyme.