# Small sample t-test for a population mean

Experience raising New Jersey Red chickens revealed the average weight of the chickens at age 5 months to be 4.35 lbs. The weights are normally distributed.

In an effort to increase their weight, a special additive was mixed with the chicken feed. The subsequent weights of a sample of five-month-old chickens were (in pounds): 4.41, 4.37, 4.33, 4.35, 4.30. Level of significance = 0.01.

1. Has the special additive increased the weight of the chickens?

2. State the hypothesis to be tested, the decision rule, the test statistic and the decision.

3. Estimate the p-value

4. In the context of your hypothesis, write out in words what making a type I error would mean.

Please provide detailed answers so that I can verify my answers and the calculations to my answers are proper/correct. Thank you.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

I have attached a Word document with detailed calculations and explanations of all calculations for performing a small sample hypothesis test (t-test) regarding a population mean.

Experience raising New Jersey Red chickens revealed the average weight of the chickens at age 5 months to be 4.35 lbs. The weights are normally distributed.

In an effort to increase their weight, a special additive was mixed with the chicken feed. The subsequent weights of a sample of five-month-old chickens were (in pounds): 4.41, 4.37, 4.33, 4.35, 4.30. Level of significance = 0.01.

1. Has the special additive increased the weight of the chickens?

2. State the hypothesis to be tested, the decision rule, the test statistic and the decision.

3. Estimate the p-value

4. In the context of your hypothesis, write out in words what making a type I error would mean.

Please provide detailed answers so that I can verify my answers and the calculations to my answers are proper/correct. Thank you.

Solution

Detailed calculations are shown below. Verbal explanations are provided here.

1. Has the special additive increased the weight of the chickens?

The test of hypothesis did not provide enough evidence to show that the population mean weight of chickens taking the special additive is greater after 5 months than the population mean weight of chickens not taking the additive. See the decision rule in the detailed calculations below.

2. State the hypothesis to be tested, the decision rule, the test statistic and the decision.

The null and alternative hypotheses are shown in the detailed calculations below. The null simply says that the mean weight of chickens taking the additive is the same as the mean weight of chickens not taking the additive. The alternative says the chickens taking the additive have mean weight higher than the mean of those not taking the additive.

3. Estimate the p-value

A detailed explanation of the p-value is provided in the calculations below.

4. In the context of your hypothesis, write out in words what making a type I error would mean.

In general, a Type I error is rejecting the null hypothesis when it is really true. In terms of this problem a Type I error would be to falsely conclude that the mean weight of chickens taking the additive is greater than the mean of those not taking it, when in fact there is no difference between the two means.

Note that the probability of making a Type I error is stated ahead of time in these types of problems and is called the "significance level" or . In this problem the probability of making a type I error is 0.01.

First calculate the following summary statistics for this problem:

(The sample size)

(The sample mean)

(The sample standard deviation)

The null and alternative hypotheses for this problem are as follows:

The decision rule:

If the null hypothesis is true, the test statistic has a t distribution with 4 degrees of freedom. We want to reject the null hypothesis if the test statistic does not appear to belong to this distribution. The level of significance tell us that we must reject the null hypothesis if the test statistic is greater than the percentile of the t distribution with 4 degrees of freedom (in this case the 99th percentile). The 99th percentile of a t distribution with 4 degrees of freedom is 3.747. So our decision rule is:

Reject the null hypothesis is the test statistic is greater than 3.747, or equivalently:

Reject if t > 3.747.

The calculated value of the test statistic is:

Note that the symbol refers to the "null value" in the hypotheses ( = 4.35)

Note also that since the test statistic does not exceed 3.747, the decision rule above tells us we should not reject the null hypothesis.

The p-value:

The p-value is defined as the probability that a randomly selected member of the t distribution with 4 degrees of freedom exceeds the observed value of the test statistic.

Some classes use t-tables to estimate this. Others use Excel functions to get exact values. I have used the FINV function in Excel and found that the exact p-value is 0.4597.

Note that another way to create a decision rule might be to say "Reject the null hypothesis if the p-value is less than , or:

Reject the null if p-value < 0.01. As with the other way to state the decision rule, we cannot reject the null hypothesis.

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