How did the professional graphologists do? Our experiment was based on a BBC stu

Question

How did the professional graphologists do? Our experiment was based on a BBC study from the 1990s, which tested the ability of professional handwriting analysts (graphologists) to do exactly what you did: classify Monk or Actor based on handwriting alone. In the BBC study, each graphologist was given not just two samples of handwriting to classify, but 30, half of which came from Monks and half from Actors. One of the graphologists (Graphologist A) correctly classified 20 of the 30 samples as either Monks or Actors. (a) We want to know whether we have evidence that Graphologist A was doing better than guessing, with 50% chance of being correct each time, when she correctly classified 20 out of 30 handwriting samples. Let X be the number of correct responses out of the 30 samples. Write down the appropriate null hypothesis, H_0, and alternative hypothesis, H_1, in terms of the distribution of X and its parameters. Use a two-sided alternative, because guesses could be either better or worse than chance. (b) Sketch as a curve the shape of the probability function of X under the null hypothesis. Your sketch should have axes labelled x and P(X = x). Mark on the sketch the upper and lower limits of x, and the approximate value of x where the curve peaks under the null hypothesis. Also mark the observed value of X so that you can see the tail probabilities required for the p-value, and shade under the curve the area represented by the p-value. (c) Write down the R command required to find the p-value for the hypothesis test, and run this command in R to find the p-value. Interpret the result in terms of strength of evidence against the null hypothesis. Do we have evidence that Graphologist A was doing better than making a 50-50 guess on the 30 handwriting samples? The real study covered not just monks and actors, but several other classes of people: entrepreneurs and librarians: good and bad secretaries: and nuns versus actresses. It also involved four professional graphologists, not just one. In total, there were 600 handwriting samples given to the professional 3 graphologists, and they correctly classified 400 between them. (d) What is the sample proportion of correct classifications from the full study? How does it compare with the sample proportion of correct classifications by Graphologist A in the Monks/Actors study? (e) Let Y be the number of correct classifications out of 600 samples in the full study. Follow all the steps in (a), (b), and (c), but using random variable Y instead of X, to decide whether or not the scores of the graphologists in the full study indicated a performance better than 50-50 guessing.

Explanation / Answer

Solution:

Part a

Here, we have to use z test for population proportion. The null and alternative hypotheses for this test are summarised as below:

Null hypothesis: H0: The probability of correct response is 50%.

Alternative hypothesis: Ha: The probability of correct response is other than 50%.

Symbolically, it is given as below:

H0: p = 0.5 versus Ha: p 0.5

This is a two tailed test.

We assume 5% level of significance or alpha value as 0.05 for this test.

The test statistic formula for this test is given as below:

Z = (P – p)/sqrt(pq/n)

Where, q = 1 – p = 1 – 0.5 = 0.5

We are given

X = number of correct responses = 20

n = 30

Sample proportion = P = X/n = 20/30 = 0.67

Z = (0.67 – 0.50) / sqrt(0.5*0.5/30)

Z = 1.862257

p-value = 0.0629

alpha value = 0.05

P-value > Alpha value

So, we do not reject the null hypothesis that the probability of correct response is 50%.

There is sufficient evidence to conclude that the probability of correct response is 50%.

Part b

Here, we have to sketch a curve and we have to show p-value associated with above test. Required curve is given as below:

Part c

R command required to find p-value for above test is given as below:

> prop.test(x=20, n=30, p=0.5, correct = FALSE)

Output:

        1-sample proportions test without continuity correction

data: 20 out of 30, null probability 0.5

X-squared = 3.3333, df = 1, p-value = 0.06789

alternative hypothesis: true p is not equal to 0.5

95 percent confidence interval:

0.4878005 0.8076950

sample estimates:

        p

0.6666667

Part d

We are given X = 400 and n = 600

Required sample proportion = X/n = 400/600 = 4/6 = 2/3 = 0.6667

Part e

Here, we are given n = 600

Y = 400

Sample proportion = Y/n = 400/600 = 0.6667

Null hypothesis: H0: The probability of correct response is 50%.

Alternative hypothesis: Ha: The probability of correct response is greater than 50%.

Symbolically, it is given as below:

H0: p = 0.5 versus Ha: p > 0.5

This is a one tailed test. This is upper tailed or right tailed test.

We assume 5% level of significance or alpha value as 0.05 for this test.

The test statistic formula for this test is given as below:

Z = (P – p)/sqrt(pq/n)

Where, q = 1 – p = 1 – 0.5 = 0.5

We are given

X = number of correct responses = 20

n = 30

Sample proportion = P = X/n = 20/30 = 0.67

Z = (0.67 – 0.50) / sqrt(0.5*0.5/30)

Z = 1.862257

p-value = 0.0339

alpha value = 0.05

P-value < Alpha value

So, we reject the null hypothesis that the probability of correct response is 50%.

There is sufficient evidence to conclude that the probability of correct response is greater than 50%.

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