SampleVarianceandStandardError: You encounter a deck of Martian playing cards.There are 87 cards in the deck. You cannot read Martian, and so the meaning of the cards is mysterious. However, you notice that some cards are blue, and others are yellow.(a) You shuffle the deck, and draw one card. You repeat this exercise 10 times, replacing the card you drew each time before shuffling. You see 7 yellow and 3 blue cards in the deck. As you know, the maximum likelihood estimate of the fraction of blue cards in the deck is 0.3. What is the standard error of this estimate?(b) How many times would you need to repeat the exercise to reduce the standard error to 0.05?

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warylucknow warylucknow · Answer 1 · 2020-03-25T13:08:38+01:00

Answer:

(a) The standard error of estimate of the fraction of blue cards in the deck is 0.145.

(b) The sample size required is 84.

Step-by-step explanation:

The proportion of blue cards is, p = 0.30.

The number of times the cards were shuffled is, n = 10.

(a)

The standard error for sample proportion is given by:

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

Compute the value of standard error as follows:

[tex]SE_{p}=\sqrt{\frac{p(1-p)}{n}}\\=\sqrt{\frac{0.30\times (1-0.30)}{10}}\\=\sqrt{0.021}\\=0.145[/tex]

Thus, the standard error of estimate of the fraction of blue cards in the deck is 0.145.

(b)

Now we need to reduce the standard error.

The standard error is inversely proportional to the sample size.

If the sample size is increased then the standard error will be decreased.

So we need to use a larger sample size.

Compute the value of n for standard error 0.05 and p = 0.30 as follows:

[tex]SE_{p}=\sqrt{\frac{p(1-p)}{n}}\\0.05=\sqrt{\frac{0.30(1-0.30)}{n}}\\0.05^{2}=\frac{0.21}{n}\\n=84[/tex]

Thus, the sample size required is 84.