Question

A juice corporation, a producer of frozen orange juice, claims that more than 20.4% of all orange-juice drinkers prefer its product. To test the validity of this claim, a market researcher took a random sample of 400 orange-juice drinkers and found that 97 preferred the juice brand. Does the sample evidence demonstrate the validity of the juice claim. (Use significance level 5%) (a) What is the value from the z-test? (b) is it significant at 5% (c) calculate the lower bound for the 95% confidence interval for the proportion of 96 people who prefer the juice brand (d) From Question 6, calculate the upper bound for the 95% confidence interval for the proportion of 96 people who prefer juice brand

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H₀ : p = 0.204

H_a : p > 0.204

n = 200

x =97

= x / n = 97 /400 = 0.242

P₀ = 0.204

1 - P₀ = 1 - 0.204 =0.796

a ) Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.242 -0.204/ [0.204 * 0.796 / 400 ]

= 1.911

Test statistic = z = 1.91

P(z > 1.91 ) = 1 - P(z < 1.91 ) = 1 - 0.9719

P-value = 0.0281

b ) = 0.05

P-value <

0.0281 < 0.05

Reject the null hypothesis .

There is sufficient evidence to suggest that  

c ) Given that

n = 400

x =96

= x / n = 96 / 400 = 0.240

1 - = 1 - 0.240 = 0.760

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.960 * (((0.240 * 0.760) / 400)

= 0.042

d ) A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.240 - 0.042 < p < 0.240 + 0.042

0.198 < p < 0.282  

The lower bound = 0.198

The upper bound = 0.282