Dataset ex1063 is available here Many consumers pay careful attention to stated
ID: 3311677 • Letter: D
Question
Dataset ex1063 is available here Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases. It is therefore important that the information on packages be accurate. A random sample of n 12 frozen dinners of certain type was selected from production during particular period, and the calorie content of each one was determined. (This determination entails destroying the product, so a census would certainly not be desirable!) Here are the resulting observations, along with a boxplot and normal probability plot. (To obtain the dataset for your analysis software, go to the Book Companion Website.) (a) Is it reasonable to test hypotheses about true average calorie content by using a t test? Yes, it is reasonable. O No, t test is not applicable here. O It depends on the results of t test (b) The stated calorie content is 241. Does the boxplot suggest that true average content differs from the stated value? O Yes, it is clear that true average content differs from the stated value. No, it is possible that true average content is 241. There's not enough evidence to decide. (c) Carry out a formal test of the hypotheses suggested in part (b). Use Table 4 in Appendix A. Use -0.05. Round your test statistic to two decimal places and your P-value to three decimal places.) Conclusion: o Reject Ho O Fail to reject Ho- You may need to use the appropriate table in Appendx A to answer this question.Explanation / Answer
a.
yes, it is reasonable to using the T test because they sample data
b.
yes, it is clear that true average content differ the stated content value because given stated value is 241(sample mean = 242.4167 from sample data)
c.
Given that,
population mean(u)=241
sample mean, x =242.4167
standard deviation, s =12.6092
number (n)=12
null, Ho: =241
alternate, H1: !=241
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.2
since our test is two-tailed
reject Ho, if to < -2.2 OR if to > 2.2
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =242.4167-241/(12.6092/sqrt(12))
to =0.39
| to | =0.39
critical value
the value of |t | with n-1 = 11 d.f is 2.2
we got |to| =0.39 & | t | =2.2
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.3892 ) = 0.7046
hence value of p0.05 < 0.7046,here we do not reject Ho
ANSWERS
---------------
null, Ho: =241
alternate, H1: !=241
test statistic: 0.39
critical value: -2.2 , 2.2
decision: do not reject Ho
p-value: 0.7046 = 0.705
fail to reject the null hypothesis
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