Chapter 9.2, Problem 46E

a)

To determine

The reason why we presented the two cups to each coffee drinker in a random order.

Explanation of Solution

Given:A hypothesis test is run to verify the claim that only half of all coffee drinkers prefer fresh brewed coffee.

To run a hypothesis test one of the required condition is to pick the sample randomly.

Here presenting the two cups to each coffee drinker in a random order, so that some people tasted the fresh coffee first, while others drank the instant coffee first. This confirms that the people have taken the cups randomly which satisfy the condition.

b)

To determine

The verify the claim that only half of all coffee drinkers prefer fresh brewed coffee.

Answer to Problem 46E

At 5% level of significance, we have sufficient evidence to reject the null hypothesis and conclude coffee drinkers prefer fresh brewed coffee.

Explanation of Solution

Given:A hypothesis test is run to verify the claim that only half of all coffee drinkers prefer fresh brewed coffee.

Sample size = n = 50

Number of people prefer fresh coffee = 36

Concept used:The conditions to satisfy to run one sample Z test

1) The sample should be random

2) np = 10

3) n(1-p) = 10

4) There should be at least 10% of people of population in the sample.

Test statistic

  $Z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

If p value is less than the level of significance, we have sufficient evidence to reject the null hypothesis.

Calculation:Let p be the proportion of people prefers fresh brewed coffee in the population.

Null Hypothesis and alternate hypothesis are shown below

  $\begin{array}{l}{H}_0:p=0.5\\ H_a:p>0.5\end{array}$

The proportion of people prefers fresh brewed coffee in sample is

  $\hat{p}=\frac{36}{50}=0.72$

  $p_0=0.5$

The sample of 50 persons selected coffee cups randomly,

  $\begin{array}{l}n{p}_0=50*0.5=25\\ n{p}_0\ge 10\\ n(1-p_0)=50*(1-0.5)=25\\ n(1-p_0)\ge 10\end{array}$

Here there will be at least 500 coffee drinkers, so 10% condition is satisfied.

So here we run a one sample Z test. Test statistic is calculated as shown below

  $\begin{array}{l}Z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\\ Z=\frac{0.72-0.50}{\sqrt{\frac{0.5(1-0.5)}{50}}}=\frac{0.22}{0.0707}=3.11\end{array}$

The p value is P(Z = 3.11)

using standard Z table we get P(Z < 3.11) = 0.9991

P(Z = 3.11) = 1 − P(Z < 3.11) = 1 − 0.9991 = 0.0009

The p value is less than level of significance 0.05, So we have sufficient evidence to reject the null hypothesis.

Conclusion:

At 5% level of significance, we have sufficient evidence that coffee drinkers favor fresh brewed coffee over instant coffee.