Question

The t statistic for a test of ?0: ?= 42 ??:? ≠ 42
based on n = 6 observations has the value t = -1.93.

(a) What are the degrees of freedom for this statistic?

(b) Using the appropriate table in your formula packet, bound the p-value as closely as possible:
< p-value <

Answer 1

(a)

This is a one sample t-test.

So, the degree of freedom = n-1,

where 'n' is the number of observations.

here n = 6, so degree of freedom = 6-1 = 5.

Hence, the degree of freedom of the t-statistic = 5.

(b)

We are given that the t-value for the test is = -1.93. And from the alternate hypothesis we can say that it is a two tailed test as we are testing just the equality of the population mean.

So, the two tailed p-value for the test statistics value of t = -1.93 at 5 degrees of freedom can be obtained from the standard t-distribution table is, 0.0557

So, 0.05 < p-value < 0.10

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