# Testing of Hypothesis

1. Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, a range of numbers for the P-value, critical value and state the final conclusion.

Claim:- The mean bottom temperature of a fish breeding pond is less than 20 degree celcius.

Sample Data:- n=56, x-bar =19.3, s=1.6. The significance level is alpha = 0.05

Calculate the test statistic, p-value using t-table, critical value using t-table and conclusion.

2. A random sample of 56 measured the accuracy of their wristwatches, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 56 values have a mean of -94.7 sec and a standard deviation of 152.6 sec. Use a 0.01 significance level to test the claim that the population of all watches has a mean equal to 0.00 sec

What is the correct conclusion?

1. There is not sufficient evidence to warrant rejection of the claim that the population mean is 0.00 sec.

2. There is sufficient evidence.

3. The birth weights are recorded for a sample of male babies born to mothers taking a special vitamin supplement. When testing the claim that the mean birth weights for all male babies of mothers given vitamins is equal to 3.39 kg, which is the mean for the population of all males. The significance level is alpha =0.05

N= 17.0000, sx = 0.6300, x bar = 3.3100

4. Find the test statistic, find the critical values of chi-square and limits containing the p-value; then determine whether there is sufficient evidence to support the given alternative hypothesis

H1:

Determine the test statistic, use chi-square table to determine the critical values and final conclusion.

5. Use the traditional method with alpha =0.05 to test the claim that a new method of laser cutting reduces the variation in lengths of 2x4 lumber. The standard deviation of the lengths of the sawed 2x4s is 0.33 inches. A simple random sample of 91 lasercut 2x4s, taken from a normally distributed population, was found to have a standard deviation of 0.3 inches.Calculate the test statistic and the critical value to determine the correct conclusion.

#### Solution Summary

Calculation of one sample t test and Chi square test.