Let’s Go Hamburger Restaurant does not want to display the nutritional contents
ID: 3321452 • 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
P = 0.5
p = 49/74 = 0.66
H0: P = 0.5
H1: P > 0.5
The test statistic Z = (p - P)/sqrt (P * (1 - P)/n)
= (0.66 - 0.5)/sqrt (0.5 * 0.5/74)
= 1.38
P-value = P(Z > 1.38)
= 1 - P(Z < 1.38)
= 1 - 0.9162 = 0.0838
As the p-value is greater than the significance level (0.0838 > 0.05), so the null hypothesis is not rejected.
So at 5% significance level there is not sufficient evidence to support that more than 50% of the people read nutrition labels.
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