Question

BACK A famous (in statistical circles) study involves a woman who claimed to be able to tell whether tea or milk was poured first into a cup. She was presented with eight cups containing a mixture of tea and milk, and she correctly identified which had been poured first for all eight cups. Is she doing better than if she were just guessing? Which of the following best defines the parameter of interest in the context of the study and uses the appropriate symbol? The number of times the lady correctly identifies which was poured first, p-hat The number of times the lady correctly identifies which was poured first, n The long-run proportion of times the lady correctly identifies which was poured first, p-hat. The long-run proportion of times the lady correctly identifies which wat poured first, n O Which of the following best describes the null hypothesis and the alternative hypothesis using the appropriate symbol? O null: π-0.5, alternative: π > 0.5 nu,m: Ri-0.5 , alternative: R> 0.5 null: 0.5, alternative: p < 0.5 O null: π 0.5, alternative: π < 0.5 cerwed.nrnnortion of times the ladv enerectlu identified what was nouured first intn the eun? What

Answer 1

Given data

Number of trail (n)=8

number of times lady identifies correctly (x)=8

1) the best define the parameter of interest in given context is

the long run proportion of the times the lady correctly identifies which was poured first p-hat

2)

here we have to check for the claim that is she is doing better than simply guessing

So the hypothesis will be

$H_{o}:\pi =0.5$

$H_{a}:\pi >0.5$

3) we have

Number of trail (n)=8

number of times lady identifies correctly (x)=8

So $\hat{p}=\frac{x}{n}=\frac{8}{8}=1$

4)The anticipation for the center of the distribution will be

0.5 because that is the value of the parameter if the null hypothesis is true .the standard deviation will be positive because the standard deviation must be at least 0 and it will be equal to 0 if there is no variability in the values .

5)

test statistic

$Z=\frac{\hat{p}-\pi }{\sqrt{\frac{\pi (1-\pi )}{n}}}=\frac{1-0.5}{\sqrt{\frac{0.5*0.5}{8}}}=2.83$

6)best description of the standardized statistic will be

the standardized   statistic represent the number of standard deviation the observed proportion is above 0.5

7)conclusion

we have strong evidence that the long run proportion of times that the lady makes the correct identification is greater than 50%