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strojnjashka [21]

3 years ago

10

a charity sold 475 tickets to a dinner auction. if the charity raised $ 16,625 in ticket sales , what was the cost of one ticket

?

1 answer:

Nuetrik [128]3 years ago

7 0

$16,625.00/475=$35.00

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Rudiy27

Answer:

The sample size required is 910.

Step-by-step explanation:

The 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 is:

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

Given:

$\hat p = 0.16\\MOE= 0.02\\Confidence\ level =0.90$

The critical value of <em>z</em> for 90% confidence level is:

$z_{\alpha /2}=z_{0.10/2}=z_{0.05}=1.645$ *Use a standard normal table.

Compute the sample size required as follows:

$MOE=z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }\\0.02=1.645\times \sqrt{\frac{0.16(1-0.16)}{n} }\\n=\frac{(1.645)^{2}\times 0.16\times (1-0.16)}{(0.02)^{2}} \\=909.2244\\\approx910$

Thus, the sample size required is 910.

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