SAT College

The blue catfish (Ictalurus Furcatus) is the largest species of North American catfish. The current world record stands at 143 pounds, which was caught in the John H. Kerr Reservoir (Bugg's Island Lake) located in Virginia. According to American Expedition, the average weight of a blue catfish is between 20 to 40 pounds. Given that the largest blue catfish ever caught was at the John H. Kerr Reservoir, you believe that the mean weight of the fish in this reservoir is greater than 40 pounds. Use the data below, which represents the summary statistics for 53 blue catfish caught at this reservoir, and a 0.05 significance level to test the claim that the mean weight of the fish in the John H. Kerr Reservoir is greater than 40 pounds.
;

pounds;
pounds

a) Identify the null and alternative hypotheses?

:
?



:
?



b) What type of hypothesis test should you conduct (left-, right-, or two-tailed)?
left-tailed
right-tailed
two-tailed
c) Identify the appropriate significance level.


d) Calculate your test statistic. Write the result below, and be sure to round your final answer to two decimal places.


e) Calculate your p-value. Write the result below, and be sure to round your final answer to four decimal places.


f) Do you reject the null hypothesis?
We reject the null hypothesis, since the p-value is less than the significance level.
We reject the null hypothesis, since the p-value is not less than the significance level.
We fail to reject the null hypothesis, since the p-value is less than the significance level.
We fail to reject the null hypothesis, since the p-value is not less than the significance level.

g) Select the statement below that best represents the conclusion that can be made.
There is sufficient evidence to warrant rejection of the claim that the mean weight of the fish in the John H. Kerr Reservoir is greater than 40 pounds.
There is not sufficient evidence to warrant rejection of the claim that the mean weight of the fish in the John H. Kerr Reservoir is greater than 40 pounds.
The sample data support the claim that the mean weight of the fish in the John H. Kerr Reservoir is greater than 40 pounds.
There is not sufficient sample evidence to support the claim that the mean weight of the fish in the John H. Kerr Reservoir is greater than 40 pounds.

Answers

Answer 1

Note that the answers to the statistical question on hypothesis are given below.

What is the null hypothesis?

a) Null hypothesis: mean weight of fish in John H. Kerr Reservoir <= 40 pounds.

Alternative hypothesis: mean weight of fish in John H. Kerr Reservoir > 40 pounds.

b) Right-tailed hypothesis test.

c) Significance level = 0.05.

d) The test statistic cannot be calculated without the sample mean, standard deviation, or sample size.

e) The p-value cannot be calculated without the sample mean, standard deviation, or sample size.

f) The test statistic and p-value cannot be calculated without the sample mean, standard deviation, or sample size.

g) None of the given statements can be selected without the necessary sample data.

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Related Questions

What is permissive mean

Answers

The term "permissive" generally means granting permission or allowing something to be done without objection or interference. It implies a degree of flexibility or leniency in terms of rules or regulations. For example, a permissive parent may allow their child to make their own decisions without strict rules or discipline. In contrast, an authoritarian parent may have strict rules and expect strict obedience from their child. In a legal context, a permissive law or regulation allows something to be done, but does not necessarily require it.

According to the Bureau of Labor Statistics, 71.9% of Young women enroll in college directly after high school graduation. Suppose a random sample of 200 female high school graduates is selected and the proportion who enroll in college is obtained.


a. What value should we expect for the sample proportion?

b. What is the standard error?

c. What effect would increasing the sample size to 500 have on the standard error?

Answers

a. The expected value for the sample proportion can be calculated as the population proportion: 71.9%, so we expect that approximately 71.9% of the 200 female high school graduates will enroll in college directly after high school graduation.

Expected value = population proportion * sample size = 0.719 * 200 = 143.8

Therefore, we expect approximately 144 female high school graduates to enroll in college directly after high school graduation.

b. The standard error of the sample proportion can be calculated as:

SE = sqrt[p(1-p)/n], where p is the population proportion and n is the sample size.

SE = sqrt[0.719(1-0.719)/200] = 0.033

Therefore, the standard error is 0.033.

c. The standard error is inversely proportional to the square root of the sample size. Increasing the sample size from 200 to 500 would reduce the standard error:

SE = sqrt[0.719(1-0.719)/500] = 0.023

Therefore, the standard error would decrease to 0.023 if the sample size increased to 500.
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