A company that manufactures a particular brand of chocolate bars is concerned that the mean weight of chocolate bars should not exceed 150 grams. A manager takes a random sample of 16 chocolate bars to determine if the manufacturing process needs any adjustment. The sample mean and the standard deviation are found to be 150.7 grams and 2.5 grams respectively. Using the computer output below, or otherwise, answer the questions that follow. Test of mu = 150 vs > 150 95% Lower N Mean SE Mean Bound t p 16 150.700 0.625 149.604 1.12 0.140.
a) Discuss why, in this case, one has to assume that the sample means are normally distributed.
b) Verify the value of the standard error of the mean.
c) Clearly state the null and the alternative hypothesis in the context of this problem?
d) State the p-value and the test-statistic from the output above.
e) Discuss, using the p-value at the 5% significance level, whether there is any evidence that suggests the population mean weight of the chocolate bars is greater than 150 grams? Give justification of your answer in context of the problem.
A company manufactures a particular brand of chocolate bars and they are concerned about the mean weight of the chocolate bars to be within 150 grams.
Number of samples chosen randomly is 16.
The mean of the chosen samples was found to be 150.7 grams and the standard deviation of the chosen samples was found to be 2.5 grams.
- According to the Central Limit theorem, the mean of a large sample, each of which have a well defined expected value and variance, follows normal distribution. This implies that the mean of any sample drawn from the large population varies around the population mean following normal distribution. Thus, according to the central limit theorem, the sample mean is assumed to be following normal distribution. While varying around the population mean, the sample mean shows normal distribution. This can be proved by drawing histograms of the parent population.
- Standard error of the mean is given by the formula = standard deviation / sqrt (number of samples) = 2.5 / sqrt (16) = 2.5 / 4 = 0.625. The value of the standard error given in the output matches the standard error found on dividing the standard deviation by the square root if number and both the values are equal to 0.625. Thus, the value of standard error is found to be correct in the output.
- The null hypothesis and the alternative hypothesis of the problem is given as follows:
H0 : mean of the sample = 150
H1 : mean of the sample is > 150
The p value of the test, as given in the output is 0.140 and the value of the test statistics from the given output was found to be 1.12.
The given level of significance is 5%. In order to test the hypothesis at 5% level of significance, the “null hypothesis” and “alternative hypothesis” was considered. The “null hypothesis” is considered that the mean value of the sample is equal to 150 grams. The “alternative hypothesis” is that the mean value of the sample is greater than 150 grams. The p value of the test was found to be 0.140 which is greater than 0.05, the considered “level of significance” of the test. The p value found from the output of the test is more than 0.05. This suggests that the test is insignificant and the “null hypothesis” is accepted. This shows that the mean weight of the chocolate bars is equal to 150 grams.