Question

# The Student's t distribution table gives critical values for the Student's t distribution. Use an appropriate...

Sample size = n = 51

Sample mean = = 18.5

Standard deviation = s = 2.7

Claim: The average length of a trout caught in Pyramid Lake is less than μ = 19 inches.

The null and alternative hypothesis is

Level of significance = 0.05

Here population standard deviation is unknown so we have to use t-test statistic.
Test statistic is

Degrees of freedom = n - 1 = 51 - 1 = 50

Critical value = 1.684    ( Using t table)

Test statistic | t | <  critical vaue we fail to reject null hypothesis.

Conclusion:  Fail to reject the null hypothesis, there is insufficient evidence that the average fish length is less than 19 inches.

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By using the P-value method:

P-value = P(T < - 1.32) = 0.0960

P-value >0.05 we fail to reject the null hypothesis.

Conclusion:

Fail to reject the null hypothesis, there is insufficient evidence that the average fish length is less than 19 inches.

Yes, both conclusions are the same.

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