A statistics student hands each of 300 classmates 2 cookies side by side on a plate. Of the 300 students, 171 choose the cookie that’s on their right hand side, and the remaining 129 choose the cookie that’s on their left. The student says, “That’s just like tossing a coin.” The student’s friend says, “No, it’s not.” Help them settle their argument by performing a one-sample z test in Problems 1 - 3.

Question

A statistics student hands each of 300 classmates 2 cookies side by side on a plate. Of the 300 students, 171 choose the cookie that’s on their right hand side, and the remaining 129 choose the cookie that’s on their left. The student says, “That’s just like tossing a coin.” The student’s friend says, “No, it’s not.” Help them settle their argument by performing a one-sample z test in Problems 1 - 3.

1. The test should be

one-tailed.
two-tailed.

Ans - two-tailed.

2. The P-value of the test is _____%. [Please be careful to enter your answer as a percent; that is, if your answer is 50% then please enter 50 in the blank; not 50%, nor 0.5, nor 1/2, etc.]

3. The test concludes:

“That’s just like tossing a coin.”
“No, it’s not.”

Ans: "No, it's not".

Can any one help me to ans question 2?

Answer 1

you know the correct answer to the problem

2 and 6 of the test?
The P-value (problem 8) is 1.78 (it is correct)

Answer 2

To CC,

2 is B,
6 is B

Answer 3

thanks :)

had achieved good luck to the course

excuse the English'm from Portugal

Answer 4

Hey :) could you please tell me how did you get the answer 1.78 ??? I want to know how to calculate it. Thanks a lot for your help!

Answer 5

Hi CC,

Thanks for the ans.
I also want to know how to calculate this value? How did u get 1.78?
Once again thanks for the ans.

Answer 6

Mean=300*0.5=150

SD=SQRT(300*0.5*(1-0.5))=8.66

Z=+/- (170.5-150)/8.66 = +/- 2.37

Table of Z
P= 2*0.0089 = 1.78%

+/- because we have two-tailed
and multiple the value of P by 2 for the same reason

Answer 7

To answer question 2, we need to perform a one-sample z test. In this case, the null hypothesis would be that the proportion of students who choose the cookie on their right hand side is 0.5 (just like tossing a fair coin), and the alternate hypothesis would be that the proportion is not 0.5.

Let's calculate the test statistic and the p-value.

Step 1: Calculate the sample proportion
The sample proportion of students who choose the cookie on their right hand side can be calculated by dividing the number of students who chose the right cookie (171) by the total number of students (300): p̂ = 171/300 = 0.57.

Step 2: Calculate the standard error
The standard error can be calculated using the formula: SE = sqrt(p̂*(1-p̂)/n), where p̂ is the sample proportion and n is the sample size.
SE = sqrt(0.57*(1-0.57)/300) ≈ 0.0265.

Step 3: Calculate the test statistic
The test statistic, z, can be calculated using the formula: z = (p̂ - p0) / SE, where p0 is the hypothesized proportion (0.5 in this case).
z = (0.57 - 0.5) / 0.0265 ≈ 2.64.

Step 4: Calculate the p-value
Since we are performing a two-tailed test, we need to find the probability of observing a test statistic as extreme as the one calculated (in either tail) assuming the null hypothesis is true. We can use a standard normal distribution table or a calculator to find the p-value.

Using a standard normal distribution table, the p-value at z = 2.64 would be approximately 0.0082 for a two-tailed test.

Converting this p-value to a percentage, we get 0.0082 * 100 = 0.82%.

Therefore, the answer to question 2 is that the p-value of the test is approximately 0.82%.