//The Student\'s t distribution table gives critical values for the Student\'s t
ID: 3269304 • Letter: #
Question
//The Student's t distribution table gives critical values for the Student's t distribution. Use an appropriate d.f. as the row header. For a right-tailed test, the column header is the value of found in the one-tail area row. For a left-tailed test, the column header is the value of found in the one-tail area row, but you must change the sign of the critical value t to t. For a two-tailed test, the column header is the value of from the two-tail area row. The critical values are the ±t values shown.
Pyramid Lake is on the Paiute Indian Reservation in Nevada. The lake is famous for cutthroat trout. Suppose a friend tells you that the average length of trout caught in Pyramid Lake is = 19 inches. However, a survey reported that of a random sample of 46 fish caught, the mean length was
x = 18.4 inches, with estimated standard deviation s = 2.7 inches. Do these data indicate that the average length of a trout caught in Pyramid Lake is less than = 19 inches? Use = 0.05. Solve the problem using the critical region method of testing (i.e., traditional method). (Round the your answers to three decimal places.)
Find the
test statistic =
critical value =
Thanks
Explanation / Answer
State the hypotheses: The null hypothesis is the hypothesis of no difference, which states that there is no difference in the the length of a trout caught in Pyramid lake and the population length of trout in the Pyramid lake. The researcher is interested to know whether average length of a trout caught in Pyramid lake is less than population length of trout caught in Pyramid lake. The hypotheses are therefore,
H0:mu=19 versus H1:mu<19
Assumptions: Randomization condition: A random sample of 46 trout fishes were caught, so the length of the fishes are likely to be mutually independent. Moreover, assume that length of trout fishes caught is normally distributed.
The assumptions and conditions are satisfied, so use Student's t model with (n-1)=(46-1)=45 degrees of freedom to do an one-sample t test for the mean.
Test statistic.
From information given,
xbar=18.4, s=2.7, n=46
t=(xbar-mu)/(s/sqrt n)
=(18.4-19)/(2.7/sqrt 46)
=-1.507
Critical t at 45 df and alpha=0.05 is -2.014 (the test is left tailed, that is why negative sign).
Rejection rule: Reject null hypothesis, if observed t>critical t. Here, -1.507 is not greater than -2.014, that is test statistic do not fall in critical region. Therefore, fail to reject null hypothesis.
Conclusion: There is insufficient sample evidence to suggest that average length of a trout caught in Pyramid lake is less than 19 inches.
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