Question

A sample of the thickness (in hundredths of an inch) of 15 pieces of Fruit Flavored Bubble Gum yielded a sample mean of 6.841 and standard deviation of 0.155. A second independent sample of the thickness of 10 pieces of Fruit Flavored Gum yielded a sample mean of 6.701 and standard deviation of 0.108. A researcher claims thinks that Fruit Flavored Bubble Gum should be thicker on average than Fruit Flavored Gum because bubble gum has more elasticity than regular gum. Assume both distributions normal and test the researchers claim at the 0.05 significance level.

a. Make the statistical decision and write the conclusion.
b. Which test statistic is appropriate for testing the researcher's claim?

Answer 1

I answered part b first before a, bec6 the answer to part b is what would aid me in answering part a

Answer:

B. The t test statistic is the appropriate test.

A. Reject the null hypothesis

Step-by-step explanation:

X1 = 6.841, sd1 = 0.155, n1 = 15

X2 = 6.701, sd2 = 0.108, n2 = 10

1. The appropriate test statistic for this question is the t test.

T = x1-x2/√sd1²/n1 + sd2²/n2

T = 6.841-6.701/√0.155²/15 + 0.108²/10

= 2.66

H0: u1 = u2

H1; u1>u2

Significance level = 0.05

Df = n1+n2-2

= 15+10-2

= 23

T critical = 1.714

2.66>1.714

So we have to reject the null hypothesis. Since we have enough evidence that supports this claim that fruit flavor bubble gum is on average, thicker than the fruit flavored gum.

Answer 2

In this exercise we have to use the knowledge of statistics to reach a conclusion about the best way to calculate what is asked:

A) Reject the null hypothesis

B) The t test statistic is the appropriate test that results in $2.66>1.714$

In the base text, some data was informed on how to carry out the question, thus analyzing these data we found that:

• $X_1 = 6.841$
• $Sd_1 = 0.155$
• $N_1 = 15$

• $X_2 = 6.701$
• $Sd_2 = 0.108$
• $N_2 = 10$

The appropriate test statistic for this question is the t test. using the formula below to calculate:

$T = \frac{(X_1-X_2)}{\sqrt{Sd_1^2/N_1} + \sqrt{Sd_2^2/N_2}}$

Substituting the values ​​previously informed, we find that:

$T = \frac{((6.841)-(6.701)}{\sqrt{(0.155)^2/15} + \sqrt{(0.108)^2/10}}\\T= 2.66$

$H_0: u_1 = u_2\\H_1; u_1>u_2$

 

Now doing another test we found that:

$Df = N_1+N_2-2\\= 15+10-2\\= 23\\T critical = 1.714\\2.66>1.714$

So we have to reject the null hypothesis. Since we have enough evidence that supports this claim that fruit flavor bubble gum is on average, thicker than the fruit flavored gum.

See more about statistics at brainly.com/question/10951564