Answer:

Null hypothesis: H0 = 0.204

Alternative hypothesis: Ha < 0.204

z = −0.328

P value = P(Z<-0.328) = 0.3714

Decision: we FAIL to REJECT the null hypothesis.

That is, there is no convincing evidence that an increased the prices on all desserts has resulted in fewer customers choosing to order dessert

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 95% confidence interval) ---- reject Null hypothesis

Z score < Z(at 95% confidence interval) ------ accept Null hypothesis

Step-by-step explanation:

Given;

n=175 represent the random sample taken

Null hypothesis: H0 = 0.204

Alternative hypothesis: Ha < 0.204

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 175

po = Null hypothesised value = 0.204

p^ = Observed proportion = 34/175 = 0.194

Substituting the values we have

z = (0.194-0.204)/√{0.204(1-0.204)/175}

z = −0.32828

z = −0.328

To determine the p value (test statistic) at 0.05 significance level, using a one tailed hypothesis.

P value = P(Z<-0.328) = 0.3714

Since z at 0.05 significance level is between -1.96 and +1.96 and the z score for the test (z = -0.328) which falls with the region bounded by Z at 0.05 significance level. And also the one-tailed hypothesis P-value is 0.3714 which is higher than 0.05. Then we can conclude that we have enough evidence to FAIL to REJECT the null hypothesis, and we can say that at 5% significance level the null hypothesis is valid.