Question

A local bottler in Hawaii wishes to ensure that an average of 16 ounces of passion fruit juice is used to fill each bottle. In order to analyze the accuracy of the bottling process, he takes a random sample of 48 bottles. The mean weight of the passion fruit juice in the sample is 15.80 ounces. Assume that the population standard deviation is 0.8 ounce. (You may find it useful to reference the appropriate table: z table or t table) a. Select the null and the alternative hypotheses to test if the bottling process is inaccurate. Ho: μ 16; HA: μ > 16 b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic

Answer 1

Given n=48

$\mu=16$

$\bar{X}=15.80$

$\sigma=0.8$

We want to test the claim that whether the mean weight of the passion fruit juice of a local bottler in Hawaii is 16 ounces.

The Hypothesis testing problem is

$H_0:\mu =16$ Vs $H_A:\mu \neq16$

The test statistic is:

$Z=\frac{\bar{X}-\mu}{\frac{\sigma }{\sqrt{n}}}$

$Z=\frac{15.80-16}{\frac{0.8 }{\sqrt{48}}}$

$Z=\frac{-0.20}{\frac{0.8 }{6.9282}}$

$Z=\frac{-0.20}{0.1155}$

$Z=-1.7316$

The p-value is

P=2*P(Z<=-1.73)

P=2*0.0418

P=0.0836

Therefore,

0.05<=p-value<=0.10

Conclusion: Since P value is greater than 0.05 we do not reject the null hypothesis.

Therefore we do not have sufficient evidence to reject the claims that the mean weight of the passion fruit juice of a local bottler in Hawaii is 16 ounces.

Hence the mean weight of the passion fruit juice of a local bottler in Hawaii is 16 ounces.