1. A cereal company is marketing a product that contains only 100 calories. You
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Question
1. A cereal company is marketing a product that contains only 100 calories. You want to test the hypothesis that in fact their product on average does has 100 calories. You go out a sample 10 packages of the product and find that the mean of your sample X--bar is 102 and the standard deviation of you sample ‘s’ is 1.
(a) State your Null hypothesis?
(b) Perform the appropriate significance test... what is the probability that the null hypothesis is true?
(c) At an alpha level of 0.05, would you accept or reject the null hypothesis?
2. (But it's related to the first question and data, so thats why it's on the same post) The company switches to a new form of sweetener and you want to know if this change corresponds to there being a different amount of calories in the product. So you go out and collect 10 of the new formulation of the product and determine that the sample mean of these new packages is 104 calories with a standard deviation ‘s’ of 4. Recall: the data you collected for the original formulation had a mean of 102 and s = 1.
a) What type of significance test would you perform between the data collected here with those collected in ‘1’?
b) Should the test be 1 tailed or two tailed?
c) Perform the test, what is the probability the null hypothesis is true?
d) At an alpha level of 0.05, would you accept or reject the null hypothesis?
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 100
Alternative hypothesis: 100
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.3162
DF = n - 1 = 10 - 1
D.F = 9
t = (x - ) / SE
t = 6.32
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 9 degrees of freedom is less than -6.32 or greater than 6.32.
Thus, the P-value = less than 0.0001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.
Reject the H0.
The probability that the null hypothesis is true is less than 0.0001.
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