Question

Here is a link to the data set collected by Dr. Hernandez. Her research team has a data set of the weight of fishes caught from both the contaminated lake and the pristine lake. Assume both of these data sets have equal variance. Perform the correct statistical test and determine the P value to test Dr. Hernandez's hypothesis. Report your answer for P rounded to four decimal places. Use formulas, functions, and the data analysis tool pack in Excel to avoid accruing rounding errors. Useful functions include =t.test(array1,array2,tails,type) to produce a P value for any type of a t test from a data set Type can be set to 1 for paired, 2 for unpaired with equal variance, or 3 for unpaired with unequal variance =chidist(x,degrees of freedom) to produce a P value from a calculated chi squared statistic ANOVA Single Factor from the data analysis tool pack to perform a 1-way ANOVA on a data set with only one factor Regression from the data analysis tool pack to perform a linear regression and the corresponding t-test on the slope of the line Fish Weight (g)

Pristine Lake Contaminated Lake

38 79.4

86.4 77.9

77.3 99.7

90.1 65.9

69.5 98.2

56.9 80.1

59.3 66.9

75.4 88.5

60.5 89.6

87.2 80.7

78.6 95.8

45.2 88.6

75.6 84.6

81.6 57.6

74.6 98.9

54.3 85.6

88.6 85

72.6 56

53.6 68

41.6 96.8

89.5 68.4

75.9 96.7

68.5 86.4

74.6 99.6

82 60

91.3 92.8

76.9 79

80.6 76.3

86.3 69.5

62 62.7

Answer 1

In this case, Independent t-Test assuming equal population variance is suitable as our aim is to compare sizes of fish found in contaminated lake and pristine lake.

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

Here, μ1​ = Mean population fish size in pristine lake

μ2 = Mean population fish size in contaminated lake

We used excel to perform t-Test: Two-sample Assuming Equal Variance. Following is the output:

From the table, we can see that p-value for two-tailed test = 0.0135

Using the P-value approach: since p = 0.0135 < 0.05, we reject null hypothesis. Therefore, there is enough evidence to claim that population mean fish size in pristine lake​ is different than population mean fish size in contaminated lake​, at the 0.05 significance level.