Question

Respond by True or False for each question

1.

A postmix beverage machine is adjusted to release a certain amount of syrup into a
chamber where it is mixed with carbonated water. A random sample of 25 beverages
was found to have a mean syrup content of 1:05 fluid ounces and a standard deviation of
0.15 fluid ounces. Assume that the syrup content of a beverage is normally distributed.
Then, to check the claim that the syrup content per beverage is about one ounce, we
use the following test
H0 : μ = 1 against H1 : μ= 1:
Based on the samples, the p-value for the test is less than 0.05.


2.

An engineer measures the weights (in kilograms) of steel pieces. They would like to
test H0 : μ = 5 against H1 : μ > 5. The weight of a steel piece is normally distributed.
They select a random sample size of n = 25 steel pieces, and compute x̄ = 6.7 and
s = 2.37. We cannot conclude that the mean weight is larger than 5 kg at a level of
significance of 1%.

Answer 1

1.

The null and alternative hypothesis is ,

$H_0:\mu=1$

H, :μ#1

The test is two-tailed test.

The test statistic is ,

$t_{stat}=\frac{\bar{X}-\mu_0}{s/\sqrt{n}}=\frac{1.05-1}{0.15/\sqrt{25}}=1.667$

Now , df=degrees of freedom =n-1=25-1=24

The p-value is ,

p-value=$P(t_{df}>|t_{stat}|)=P(t_{24}>1.667)=0.1085$ ; The Excel function is , =TDIST(1.667,24,2)

Decision : Here , p-value > 0.05

Therefore , fail to reject Ho.

Therefore ,

Based on the samples, the p-value for the test is less than 0.05.
The above statement is false.

2.

The null and alternative hypothesis is ,

$H_0:\mu=5$

$H_a:\mu>5$

The test is right-tailed test.

The test statistic is ,

$t_{stat}=\frac{\bar{X}-\mu_0}{s/\sqrt{n}}=\frac{6.7-5}{2.37/\sqrt{25}}=3.586$

Now , df=degrees of freedom =n-1=25-1=24

The p-value is ,

p-value=$P(t_{df}>|t_{stat}|)=P(t_{24}>3.586)=0.0000$ ; The Excel function is , =TDIST(1.667,24,2)

Decision : Here , p-value <0.01

Therefore , reject Ho.

Conclusion : We can conclude that the mean weight is larger than 5 kg at a level of
significance of 1%.

Therefore ,

We cannot conclude that the mean weight is larger than 5 kg at a level of
significance of 1%.

The above statement if false.