Let’s Go Hamburger Restaurant does not want to display the nutritional contents
ID: 3301945 • Letter: L
Question
Let’s Go Hamburger Restaurant does not want to display the nutritional contents of its offerings, claiming that only 50% of its customers read them. A consumer group surveyed a random sample of 74 people and asked whether they read nutritional labels. Of this group, 49 said that they do read labels. Test the claim that more than 50% of the people read nutritional labels. Use a 5% level of significance. a. List the null and alternate hypotheses in terms of the appropriate population parameter using =, <, or >. b. Find the critical value(s) which determine(s) the rejection and acceptance regions. c. Compute the relevant test statistic and the P-value. d. Draw a conclusion. Do you reject the null hypothesis or not?
Explanation / Answer
Answer to the question)
Claim: only 50% of its customers read them
Null hypothesis: p = 0.50
Alternate hypothesis: p > 0.50
[right tailed test]
.
n = 74
x = 49
sample proportion p^ = 49 /74 = 0.66
.
SE = sqrt(p*(1-p)/n)
SE = sqrt(0.5*0.5/74)
SE = 0.0581
.
Significance level (alpha) = 0.05
.
Formula of test statistic is:
z = (p^ - p) / SE
.
on plugging the values we get
z = (0.66 -0.50) / 0.0581
z = 2.75
.
P(z > 2.75) = 1 - 0.9970 ...........[because it is a right tailed test]
P value = P(z > 2.75) = 0.003
.
Inference: Since the P value 0.003 < Alpha 0.05 , we Reject the null hypothesis
.
Conclusion: Thus we conclude that the claim that only 50% of its customers read them is incorrect. We get to know that more than 50% of the costomers read them
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