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uranmaximum [27]
3 years ago
13

A matched pairs experiment compares the taste of instant versus fresh-brewed coffee. Each subject tastes two unmarked cups of co

ffee, one of each type, in random order and states which he or she prefers. Of the 50 subjects who participate in the study, 16 prefer the instant coffee. Let p be the probability that a randomly chosen subject prefers fresh-brewed coffee to instant coffee. (In practical terms, p is the proportion of the population who prefer fresh-brewed coffee.)
Test the claim that a majority of people prefer the taste of fresh-brewed coffee. Report the large-sample z statistic and its P-value.
Mathematics
1 answer:
Bogdan [553]3 years ago
4 0

Answer:

We conclude that the majority of people prefer the taste of fresh-brewed coffee.

Step-by-step explanation:

We are given that of the 50 subjects who participate in the study, 16 prefer the instant coffee.

We have to test the claim that a majority of people prefer the taste of fresh-brewed coffee.

<em />

<u><em>Let p = proportion of the population who prefer fresh-brewed coffee</em></u>

So, Null Hypothesis, H_0 : p \leq 50%     {means that the majority of people does not prefer the taste of fresh-brewed coffee}

Alternate Hypothesis, H_A : p > 50%     {means that the majority of people prefer the taste of fresh-brewed coffee}

The test statistics that would be used here <u>One-sample z proportion statistics;</u>

                        T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion who prefer fresh-brewed coffee = \frac{34}{50} = 0.68

            n = sample of subjects = 50

So, <u><em>test statistics</em></u>  =  \frac{0.68-0.50}{\sqrt{\frac{0.68(1-0.68)}{50} } }  

                               =  2.73

The value of z test statistics is 2.73.

<u>Now, P-value of the test statistics is given by;</u>

         P-value = P(Z > 2.73) = 1 - P(Z \leq 2.73)

                       = 1 - 0.99683 = 0.0032

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of 1.645 for right-tailed test.

<em>Since our test statistics is more than the critical value of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the majority of people prefer the taste of fresh-brewed coffee.

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