Question

A random sample of 172 students was asked to rate on a scale to from 1 (not important) to 5 (extremely important) health benefits as a job characteristic (note that the rating scale can also have decimals, i.e. a student can give a rating of 1.32). The sample mean rating was 3.31, and the sample standard deviation was 0.70. For a type I error of 1% (alpha), can you be reasonably certain that the average rating is more than 3 in the population?

1)Specify the rejection region for  = 0.01. Reject H0 if

A) z > 2.33

B) t > 2.32

C) z < 2.33

D) t < 2.32

and what is your conclusion.

A soft drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of soft drink. A random sample of 49 bottles is selected, and the contents are measured. The sample yielded a mean content of 11.88 ounces with a standard deviation of 0.35 ounces.

2) State the null and alternative hypotheses.

A) H0: µ = 0, Ha: µ > 11.88

B) H0: µ = 0, Ha: µ ≠ 11.88

C) H0: µ = 0, Ha: µ > 12

D) H0: µ = 0, Ha: µ ≠ 12

Answer 1

1) Since, we are going to test for the mean rating and the alternative hypothesis would be if average rating was more than 3. Therefore, this is a right-tailed test. Also, we do not know the population variance and we use the sample variance. Hence, in this test, the test statistic would be a t-statistic.

The correct option for the rejection region can therefore be

B) t > 2.32

We can conclude that we have enough evidence to reject the null hypothesis that average rating is 3 at 1% level of significance.

2) Here denotes the average amount of soft drink the machine fills the bottles with. Now we want to test whether the amount of liquid that gets filled is not equal to 12 ounces.

Therefore the null and alternative hypothesis would be.

D)

Perform the one sample t-test in this case.

The t-test statistic is

where is sample mean,

is the hypothesised mean,

is the sample standard deviation,

n is the sample size.

t = -2.4

This is a two-tailed test and

= -2.0106

Since,

t <  

we can conclude that we reject the null hypothesis at 0.05 level of significance.