Question

Use the following to answer the questions 1-5. A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 441 gram setting. Based on a 19 bag sample where the mean is 448 grams and the variance is 625, is there sufficient evidence at the 0.05 level that the bags are underfilled or overfilled? Assume the population distribution is approximately normal. 1. What are the null and alternative hypotheses? A-Ho: ? 441, Ha: ? < 441 0: ? 441, Ha : ? 441 Ho: ? 448,Ha: ? > 448 D. Ho: ?# 441 ,Ha: ? 441 H0.12 448, Ha: ? < 448 What is the calculated value of the test statistic? A. 0.35 B. 3.84 2. D. 2.31 E. 1.80 3. Specify if the test is two-tailed, right-tailed or left-tailed. A. Two-Tailed Test B. Right-Tailed Test C. Left-Tailed Test D. None Specify the decision rule : Reject Ho if ltl> A. 0.63 B. .285 C. 1.960 D. 2.101 4. E-2585 5. What is the conclusion? C. Neither A. Reject Null Hypothesis B. Fail to Reject Null Hypothesis

Answer 1

let X1,...,Xn denote the weight of the bags of a random sample of size n

assumption is Xi~N(mu,sigma²) independently.

with mu and sigma both unknown

we want to test whether the filling machine works correctly at 441 gram setting or there is any sign of underfilled or overfilled. hence the test is both sided

1. the null hypothesis and the alternative hypothesis are

H0: mu=441 vs Ha: mu $\neq$ 441 [option B]

2. let Xbar be the sample mean. then Xbar~N(mu,sigma²/n)

but since sigma² is unknown, it is estimated by s², the sample variance

so the test statistic is T=(Xbar-441)*sqrt(n)/s which under H0 follows a t distribution with df n-1

now from the given information Xbar=448 n=19 s=sqrt(625)=25

hence value of test statistic T=(448-441)*sqrt(19)/25=1.220492 [option C]

3 since the alternative hypothesis is not equal to type hence the test is two tailed test [option A]

H0 will be rejected if the test statistic value is too low or too high

4. level of significance=alpha=0.05

since the test is two tailed,

hence reject H0 if |t|>t_alpha/2;n-1 where |t| is the absolute value of observed value of the test statistic

t_alpha/2;n-1 is the upper alpha/2 point of a t distribution with df n-1

so t_0.025;18=2.100922=2.101 [option D]

5. so |t|=1.22<t_0.025;18

hence there is not enough evidence to reject the null hypothesis.

hence the conclusion is

fail to reject null hypothesis [option B]