3. [30 points] President Food wishes to compare the weight gain of infants using
ID: 3350783 • Letter: 3
Question
3. [30 points] President Food wishes to compare the weight gain of infants using formula versus its competitor's. A sample of 100 babies using the the first its brand President Food products revealed a mean weight gain of 7.6 pounds in three months after birth. For the President Food, the sample standard deviation of the sample is 2.3 pounds. A sample of 100 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds. The sample standard deviation is 2.9 pounds. At the 05 significance level, can we conclude that babies using the President Food brand gained less weigh? a. What is the null hypothesis? b. What is the alternative hypothesis? c. State the decision rule. d. Compute the value of the test statistic. e. What is your decision regarding the null hypothesis? f. Compute the p-value and interpret it.Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
a) Null hypothesis: 1> 2
b) Alternative hypothesis: 1 < 2
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 offreedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.37014
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = 198
c) tcritical = - 1.65
We will reject the null hypothesis if t-value is less than - 1.65.
d)
t = [ (x1 - x2) - d ] / SE
t = - 1.34
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 (10) produced a t statistic of - 1.34. We use the t Distribution Calculator to find P(t < - 1.34)
f) Therefore, the P-value in this analysis is 0.091
e) Interpret results. Since the P-value (0.091) is greater than the significance level (0.05), we cannot reject the null hypothesis.
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