Question

According to Brad, consumers claim to prefer the brand-name products better than the generics, but they can't even tell which is which. To test his theory, Brad gives each of 199 consumers two potato chips - one generic, and one brand-name - then asks them which one is the brand-name chip. 92 of the subjects correctly identified the brand-name chip.
Required:
a. At the 0.01 level of significance, is this significantly greater than the 50% that could be expected simply by chance?
b. Find the test statistic value.

Answer 1

Answer:

a. There is not enough evidence to support the claim that the proportion that correctly identifies the chip is significantly smaller than 50%.

b. Test statistic z=-1.001

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that the proportion that correctly identifies the chip is significantly smaller than 50%.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.5\\\\H_a:\pi$

The significance level is 0.01.

The sample has a size n=199.

The sample proportion is p=0.462.

$p=X/n=92/199=0.462$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5*0.5}{199}}\\\\\\ \sigma_p=\sqrt{0.001256}=0.035$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.462-0.5+0.5/199}{0.035}=\dfrac{-0.035}{0.035}=-1.001$

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

$\text{P-value}=P(z$

As the P-value (0.16) is greater than the significance level (0.01), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion that correctly identifies the chip is significantly smaller than 50%.