In: Statistics and Probability
Since no data is available for question 7, hence answering only question 8 and 9.
8) To answer the given question, we construct our null and alternative hypothesis as H0: mu = 16 and H1: mu <16, where mu is the unknown population mean weight of the boxes. The test statistic for testing the given hypotheses is
t= (xbar - mu0)/(s/sqrt(n)) ; where xbar = sample mean weight, mu0 = hypothesized value of the population mean, s = sample standard deviation, n = no. of observations in the sample.
We reject H0, if t(observed) < -t(alpha, (n-1)), where t(alpha, (n-1)) is the upper alpha point of a Student's t distribution with n-1 degrees of freedom, or if the p-value is less than the level of significance (alpha).
Here t(observed) = -2.3383 and -t(alpha,(n-1)) = -2.1318 and the p-value = 0.0398 (rounded to 4-decimal place)
9) Since t(observed) < -t(alpha,(n-1)) and p-value = 0.0398 < 0.05 (level of significance, alpha), so we reject H0 and say on the basis of the given sample at 5% level of significance that there is sufficient evidence to conclude that the mean weight is actually less than 16 ounces. (Option (1) is correct).
(The answers are obtained using R-software. Code and output are attached below for verification).