Question

Agricultural scientists are testing a new chicken feed to see whether it increases the number of eggs laid. The scientists divided a flock in half and gave half of the chickens the new feed, while the other half were given a regular feed. The following table shows the eggs laid in a 1-year period for a random sample of 10 chickens using the new feed and 10 chickens using the regular feed. Assume that the population variance of the number of eggs laid per year is the same for both groups, and that the number of eggs laid per year is normally distributed. Let the chickens given the new feed be the first sample, and let the chickens given the regular feed be the second sample. At the 0.10 level of significance, is there evidence that the new feed increases the number of eggs laid? Perform the test using a TI-83, TI-83 Plus, or TI-84 calculator. Find the test statistic, rounded to two decimal places, and the p-value, rounded to three decimal places.

News Feed	Regular Feed
253	248
226	226

231	240
259	228
243	243
247	235
241	241
254	238
244	249
238	251

T & p - value = ?

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁< u₂
Alternative hypothesis: u₁ > u₂

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 4.2128
DF = 18

t = [ (x₁ - x₂) - d ] / SE

t = 0.878

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 0.878

Therefore, the P-value in this analysis is 0.196.

Interpret results. Since the P-value (0.196) is greater than the significance level (0.10), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the new feed increases the number of eggs laid.