Question

In a restaurant, the proportion of people who order coffee with their dinner is p. A simple random sample of 144 patrons of the restaurant is taken and 120 ordered coffees with their meal. (a) State the null and the alternative hypotheses if you want to test if p is greater than or equal to 0.85. (b) Determine the test statistic and test the hypotheses at 95% confidence.

Answer 1

Answer with explanation:

Given : In a restaurant, the proportion of people who order coffee with their dinner is p.

Sample size : n= 144

x= 120

$\hat{p}=\dfrac{x}{n}=\dfrac{120}{144}=0.83333333\approx0.8333$

The null and the alternative hypotheses if you want to test if p is greater than or equal to 0.85 will be :-

Null hypothesis : $H_0: p\geq0.85$   [ it takes equality (=, ≤, ≥) ]

Alternative hypothesis : $H_1: p$  [its exactly opposite of null hypothesis]

∵Alternative hypothesis is left tailed, so the test is a left tailed test.

Test statistic : $z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

$z=\dfrac{0.83-0.85}{\sqrt{\dfrac{0.85(1-0.85)}{144}}}\\\\=-0.561232257678\approx-0.56$

Using z-vale table ,

Critical value for 0.05 significance ( left-tailed test)=-1.645

Since the calculated value of test statistic is greater than the critical value , so we failed to reject the null hypothesis.

Conclusion : We have enough evidence to support the claim that p is greater than or equal to 0.85.