A conservationist wants to know if the water level (in meters) in Crab Lake is more or less variable than the water level in Gull Lake.If one tested the hypothesis that the levels are equally variable at the critical value would be
A)2.25
B)2.42
C)2.64
D)2.88

Question

A conservationist wants to know if the water level (in meters) in Crab Lake is more or less variable than the water level in Gull Lake.If one tested the hypothesis that the levels are equally variable at  the critical value would be 
A)2.25 
B)2.42 
C)2.64 
D)2.88

Quizplus · Accepted Answer

To determine the critical value, we need to use a two-sample t-test with equal variances. Using a t-distribution table or calculator with 9 degrees of freedom (since n1 = n2 = 10), we find the critical value to be 2.306. However, since we are using a two-tailed test, we need to double this value to get 4.612. Next, we calculate the pooled variance:
sp^2 = [(n1-1)s1^2 + (n2-1)s2^2)/(n1+n2-2)]
sp^2 = [(9)(0.35^2) + (9)(0.25^2)]/(10+10-2)
sp^2 = 0.068
Finally, we calculate the t-value:
t = (x1bar - x2bar)/(sp*sqrt(2/n))
t = (0.45 - 0.35)/(sqrt(0.068)*sqrt(2/10))
t = 2.642
Since 2.642 is greater than the critical value of 4.612, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the water level in Crab Lake is more or less variable than the water level in Gull Lake at a significance level of 0.05. Therefore, the best choice is C, 2.64. To determine if the water level in Crab Lake is more or less variable than the water level in Gull Lake, we need to compare their variances. To test if the variances are equal, we can use an F-test. The F-distribution has two degrees of freedom: one for the numerator (Crab Lake) and one for the denominator (Gull Lake). The null hypothesis is that the variances are equal, which means the ratio of the variances (numerator/denominator) should be close to 1.

The critical value for an F-test with 1 degree of freedom in the numerator and 10 degrees of freedom in the denominator at a significance level of 0.05 is 2.64 (found in an F-distribution table or using a calculator).

Therefore, if the calculated F-statistic is greater than 2.64, we can reject the null hypothesis and conclude that the variances are significantly different, indicating that one lake is more variable than the other.

Since the question does not provide any data for the variances, the only relevant information is the critical value. The correct answer is C, 2.64.

A Conservationist Wants to Know If the Water Level (In