Question

According to a report by the CDC, 32.6% of adults consumed fruit two or more times per day. A random sample of 500 individuals showed that 186 adults consumed fruits two or more times per day. If appropriate, test whether the population proportion of fruit consumed two or more times a day has increased, using a level of significance of α = 0.05.

a.	p-value = 0.014, reject H0.
b.	p-value = 0.0, reject H0.
c.	p-value = 0.028, reject H0.
d.	p-value = 0.72, do not reject H0.

Answer 1

Solution :

Given that,

= 0.326

1 - = 0.674

n = 500

x = 186

Level of significance = = 0.05

Point estimate = sample proportion = = x / n = 0.372

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: p = 0.326

Ha: p 0.326

Test statistics

z = ( - ) / *(1-) / n

= ( 0.372 - 0.326) / (0.326*0.674) / 500

= 2.194

P-value = P(Z >z )

= 1 - P(Z < 2.194 )

= 0.014

The p-value is p = 0.014, and since p = 0.014 < 0.05, it is concluded that the null hypothesis is rejected.

P-value = 0.014, reject H0.

Option a) is correct.