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Every day Ahmet buys a scratch-off lottery ticket with a 40%, percent chance of winning some prize. He noticed that whenever he wears his red shirt he usually wins. He decided to keep track of his winnings while wearing the shirt and found that he won 3 out of 3 times. Let's test the hypothesis that Ahmet's chance of winning while wearing the shirt is 40%, percent as always versus the alternative that the chance is somehow greater.

Sagot :

Testing the hypothesis, it is found that the p-value of the test is 0.017 < 0.05, which means that we can conclude that the red shirt increases his chances of winning.

At the null hypothesis, we test if the proportion is of 40%, that is:

[tex]H_0: p = 0.4[/tex]

At the alternative hypothesis, we test if the proportion is greater than 40%, that is:

[tex]H_1: p > 0.4[/tex]

The test statistic is:

[tex]Z = \frac{X - p}{s}[/tex]

In which:

  • X is the sample proportion.
  • p is the value tested at the null hypothesis.
  • s is the standard error, given by:

[tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex]

With n as the sample size.

In this problem:

  • 0.4 is tested at the null hypothesis, thus [tex]p = 0.4[/tex].
  • Won 3 out of 3, thus [tex]n = 3, X = \frac{3}{3} = 1[/tex].
  • The standard error is:

[tex]s = \sqrt{\frac{0.4(0.6)}{3}} = 0.283[/tex]

The value of the test statistic is:

[tex]Z = \frac{X - p}{s}[/tex]

[tex]Z = \frac{1 - 0.4}{0.283}[/tex]

[tex]Z = 2.12[/tex]

The p-value is the probability of finding a sample proportion of 1 or "above", which is 1 subtracted by the p-value of Z = 2.12.

  • Looking at the z-table, Z = 2.12 has a p-value of 0.983.
  • 1 - 0.983 = 0.017.

The p-value of the test is 0.017 < 0.05, which means that we can conclude that the red shirt increases his chances of winning.

A similar problem is given at https://brainly.com/question/24166849

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