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.

---------------------------------------------------------------------------------------

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.

#### Earn Coins

Coins can be redeemed for fabulous gifts.

##### Need Online Homework Help?

Most questions answered within 1 hours.

##### Active Questions
• V Sophia Johnson is 24 years old and single, lives in an apartment, and has no...
• CS 203 Discrete Structure 2 Create a Microsoft Access Database consisting of the following two tables:...
• Interest earned on the cash balance in the bank is recorded by the bank as: a....
• Describe the cellular mechanism of : 1- How glucose molecules are transported by the epithelial cells...
• Assume a firm is evaluating a stream of cash flows. Today the firm incurs a cost...
• In 100 words what do these measures tell you about the “spread” of the data? weight(x)...
• Consider an investment that pays $18.8 in year 1, and then stabilizes and pays$6.18 every...