Question

Neocitrus corporation, a producer of frozen orange juice, claims that more than 20% 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 80 preferred the Neocitrus brand.

(a) Does the sample evidence demonstrate the validity of the Neocitrus claim. Use significance level 5%

a. State the hypotheses.

b. State what kind of test you will be using: t-test or z-test.

c. The test value.

d. The critical value or p-value

e. The decision

(b)   (EXTRA CREDIT) Find 95% confidence interval for the proportion of people who prefer Neocitrus brand.

Answer 1

a.

p: Proportion of orange-juice drinkers who prefer Neocitrus brand

Claim : more than 20% of all orange-juice drinkers prefer Neocitrus brand : p >0.20

Null hypothesis : H_o p = 0.20

Alternative hypothesis : H_a : p >0.20

Right tailed test:

b.

kind of test you will be using z-test.: Test for single population proportion.

c.

Number of orange-juice drinkers in the sample : n=400

Number orange-juice drinkers in the sample who preferred Neocitrus brand : x =80

Sample proportion of orange-juice drinkers in the sample who preferred Neocitrus brand : = 80/400 =0.20

Hypothesized proportion : p =0.20

d.

For right tailed test :

e.

As P-Value i.e. is greater than Level of significance i.e (P-value:0.5 > 0.05:Level of significance); Fail to Reject Null Hypothesis
There is not sufficient evident to conclude that more than 20% of all orange-juice drinkers prefer Neocitrus brand

(b)

Formula for confidence interval for single population proportions : p

for 95% confidence level = (100-95)/100 =0.05

/2 =0.025

Z_/2 = Z_0.025= 1.96

95% confidence interval for the proportion of people who prefer Neocitrus brand.

95% confidence interval for the proportion of people who prefer Neocitrus brand = (0.1608,0.2392)