Question

A new soft drink is being market tested. A sample of 400 individuals participated in the taste test and 80 indicated they like the taste. At 95% confidence, test to determine if at least 22% of the population will like the new soft drink.
Required:

Determine the p-value.

Answer 1

Answer: p-value of the test  = 0.167

Step-by-step explanation:

Given that,

sample size n = 400

sample success X = 80

confidence = 95%

significance level = 1 - (95/100) = 0.05

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p = 0.22

Hₐ : p < 0.22

P = x/n = 80/400 = 0.2

Standard deviation of proportion α = √{  (p ( 1 - p ) / n }

α = √ { ( 0.22 ( 1 - 0.22 ) / 400 }

α = √ { 0.1716 / 400 }

α = √0.000429

α = 0.0207

Test statistic

z = (p - p₀) / α

z = ( 0.2 - 0.22 ) / 0.0207

z = - 0.02 / 0.0207

z = - 0.9661

fail to reject null hypothesis.

P-value Approach

P-value = 0.167

As P-value >= 0.05, fail to reject null hypothesis.

Since test is left tailed so p-value of the test is 0.167. Since p-value is greater than 0.05 so we fail to reject the null hypothesis.