Question

Hypothesis Testing 02 (when a is unkn 1) The sodium content of thirty 300 gram boxes of organic corn flakes was determined. The data (in milligram are as follows: 131.15 130.72 128.71 129.78 129.64 130.69 128.33 129.00 130.92 128.77 130.91 128.24 129.39 131.15 130.72 129.54 129.65 130.42 130.69 128.33 129.64 130.4 129.53 130.91 128.24 128.77 129.29 130.12 129.54 129.65 a. Write down the null and alternative hypotheses to test if the mean sodium content of this brand of cornflakes is 130 milligrams. b. What is the type of statistical test that that should be used? Explain c. Test the hypotheses at a 0.05, using the critical value approach. What is your decision on null hypothesis (HO)? Interpret the test results. d. What is the P Value for the test? Test the hypotheses using the P Value. Do you get the same answer as part (c)? e Construct a 95% confidence interval for the mean sodium content. Test the hypotheses using the confidence interval approach. Do you get the same answer as parts (c) and (d)?

Answer 1

mean() = 129.76, s = 0.9343

Ho: = 130 Hl: u + 130 b.) Two tailed t test

Test statistic(t) = T- 129.76 - 130 s/n E=; 0.9343/30

Degree of freedom = n-1=30-1=29

Critical value(t_a/2)=+/- 2.05

Decision Rule:

Reject the null hypothesis if: test statistic < Critical value

Here, Test statistic(-1.41)> critical value (-2.05), Fail to reject null hypothesis.

we can say that the mean sodium content of this brand of cornflakes is 130 milligram.

d.)

p-value=P(t<|-1.41|)

=0.1692

p-value > significance(alpha=0.05), fail to reject null hypothesis.

Hence,

Yes, we get same answer as in part c.

e.)

Confidence interval:

i Ita/2 * = vn 0.9343 = 129.76 = 2.05* V30 = 129.76 + 0.3497 = (129.41, 130.11)

we can see in the above confidence interval that this interval contains the true mean value 130, hence we fail to reject the null hypothesis.