Question

At Ajax Spring Water, a half-liter bottle of soft drink is supposed to contain a mean of 519 ml. The filling process follows a normal distribution with a known process standard deviation of 6 ml.
1) The normal distribution should be used for the sample mean because:_____.
a) the sample population has a large mean.
b) the population distribution is known to be normal.
c) the population standard deviation is known.
d) the standard deviation is very small.
2) Set up hypotheses and a two-tailed decision rule for the correct mean using the 5 percent level of significance. The hypothesis for a two-tailed decision is:_______.
A. H0: mu not equal to 519, H1: mu = 519, reject if z < -1.96 or z > 1.96.
B. H0: mu not equal to 519, H1: mu = 519, reject if z > 1.96 or z < -1.96.
C. H0: mu = 519, mu not equal to 519, reject if z> 1.96 or z< -1.96.
D. H0: mu = 519, H_1: mu not equal to 519, reject if z > -1.96 or z< 1.96.
a. a.
b. b.
c. c.
d. d.
3) If a sample of 16 bottles shows a mean fill of 522 ml, does this contradict the hypothesis that the true mean is 519 ml?
A) Yes.
B) No

Answer 1

Answer:

1) The normal distribution should be used for the sample mean because the population distribution is known to be normal (answer b).

2) C. H0: mu = 519, H_1: mu not equal to 519, reject if z> 1.96 or z< -1.96.

3) Yes. There is enough evidence to support the claim that the true mean is not 519 ml.

Step-by-step explanation:

1) When the population follows a normal distribution, it is correct to assume a normal distribution for the sample mean.

2) As it is a two-tailed decision rule, we are interested in detecting a significant difference below and above the mean. This is why we use the unequal sign in the alternative hypothesis.

The null hypothesis state that there is not significant difference from 519.

The critical value for a significance level of 5% is z=1.96.

$H_0: \mu=519\\\\H_a:\mu\neq 519$

3) The claim is that the true mean is not 519 ml.

Then, the null and alternative hypothesis are:

$H_0: \mu=519\\\\H_a:\mu\neq 519$

The significance level is 0.05.

The sample has a size n=16.

The sample mean is M=522.

The standard deviation of the population is known and has a value of σ=6.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{16}}=1.5$

Then, we can calculate the z-statistic as:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{522-519}{1.5}=\dfrac{3}{1.5}=2$

This test is a two-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=2\cdot P(z>2)=0.046$

As the P-value (0.046) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the true mean is not 519 ml.