Question

In the past, the population proportion of consumers preferring orange juice with no pulp was equation. To test if this population proportion has changed, a random sample of equation consumers is selected. Among the consumers in the sample, x=52 of them indicated that they prefer orange juice with no pulp. Test if the population proportion has changed at a significance level α=0.10
a. The null and alternative hypotheses should be:_______
b. Using the action limits to set up the decision rule, the decision rule should be:_____
c. The test statistic of this test is:_____
d. The conclusion of this test should be:_____

Answer 1

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

a) For the null hypothesis,

P = 0.45

For the alternative hypothesis,

P ≠ 0.45

b) For the decision rule, we would reject H0 if 0.1 > p value and accepth H0 if 0.1 < p value

Considering the population proportion, probability of success, p = 0.45

q = probability of failure = 1 - p

q = 1 - 0.45 = 0.55

Considering the sample,

Sample proportion, p = x/n

Where

x = number of success = 52

n = number of samples = 120

p = 52/120 = 0.43

c) We would determine the test statistic which is the z score

z = (p - P)/√pq/n

z = (0.43 - 0.45)/√(0.45 × 0.55)/120 = - 0.44

Recall, population proportion, P = 0.45

The difference between sample proportion and population proportion(P - p) is 0.45 - 0.43 = 0.02

Since the curve is symmetrical and it is a two tailed test, the p for the left tail is 0.45 - 0.02 = 0.43

the p for the right tail is 0.45 + 0.02 = 0.47

These proportions are lower and higher than the null proportion. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area below the test z score in the left tail 0.3336

We would double this area to include the area in the right tail of z = 0.44 Thus

p = 0.3336 × 2 = 0.67

d) Since alpha, 0.1 < than the p value, 0.67, then we would fail to reject the null hypothesis. Therefore, at a 1% level of significance, we do not have enough evidence to reject the​ null hypothesis