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last year, a soft drink manufacturer had 21% of the market. in order to increase their portion of the market, the manufacturer has introduced a new flavor in their soft drinks. a sample of 400 individuals participated in the taste test and 120 indicated that they like the taste. we are interested in determining if more than 21% of the population will like the new soft drink at the significance level 0.1. round your solutions for this exhibit to 4 decimal places. calculate the p-value for the test.

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At the significance level of 0.1, there is sufficient evidence to conclude that more than 21% of the population will like the new soft drink.

State the hypotheses

Null hypothesis (H0): The proportion of the population that likes the new soft drink is equal to or less than 21%.

Alternative hypothesis (Ha): The proportion of the population that likes the new soft drink is greater than 21%.

Calculate the sample proportion

Sample proportion (
\bar p) = 120/400 = 0.3

Calculate the standard error

Standard error (SE) = √[(
\bar p(1 -
\bar p)/n] = √[0.3(1-0.3)/400] = 0.033

Calculate the z-score

z-score = (
\bar p- p) / SE = (0.3 - 0.21) / 0.033 = 2.73

Find the p-value

The p-value is the probability of getting a z-score as extreme or more extreme than 2.73, assuming the null hypothesis is true. We can find the p-value using a z-table or a calculator. The p-value is approximately 0.0034.

Make a decision

Since the p-value (0.0034) is less than the significance level (0.1), we reject the null hypothesis.

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