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Marina CMI [18]
3 years ago
15

How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l

ength less than 0.1, 0.05, and 0 .01, respectively
Mathematics
1 answer:
musickatia [10]3 years ago
8 0

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The margin of error in this interval is:

MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The formula to compute the sample size is:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}

(1)

Given:

\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97

Thus, the minimum sample size required is 97.

(2)

Given:

\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385

Thus, the minimum sample size required is 385.

(3)

Given:

\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604

Thus, the minimum sample size required is 9604.

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