The Student\'s t distribution table gives critical values for the Student\'s t d

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 alpha found in the one-tail area row. For a left-tailed test, the column header is the value of alpha found in the one-tall 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 alpha from the two-tail area row. The critical values are the plusorminus values shown. A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.13 years, with sample standard deviation s = 0.77 years. However, it is thought that overall population mean age of coyotes is mu = 1.75. Do the sample data indicate that coyotes in this region Minnesota tend to live longer than the average of 1.75 years? Use alpha = 0.001. Solve the problem using the critical region method of testing (i.e., traditional method). (Round your answers to three decimal places.) test statistic = critical value =

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: < 1.75

Alternative hypothesis: > 1.75

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.1135

DF = n - 1 = 46 - 1

D.F = 45

t = (x - ) / SE

t = 3.35

tcritical = 2.412

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 3.35.

Thus the P-value in this analysis is 0.00082.

Interpret results. Since the P-value (0.00082) is less than the significance level (0.01), we have to reject the null hypothesis.

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