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sweet-ann [11.9K]
3 years ago
14

A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. A previous study indicates that the propo

rtion of left-handed golfers is 8%. How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?
Mathematics
1 answer:
never [62]3 years ago
3 0

Answer:

A sample of 997 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

A previous study indicates that the proportion of left-handed golfers is 8%.

This means that \pi = 0.08

98% confidence level

So \alpha = 0.02, z is the value of Z that has a p-value of 1 - \frac{0.02}{2} = 0.99, so Z = 2.327.  

How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

This is n for which M = 0.02. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.02 = 2.327\sqrt{\frac{0.08*0.92}{n}}

0.02\sqrt{n} = 2.327\sqrt{0.08*0.92}

\sqrt{n} = \frac{2.327\sqrt{0.08*0.92}}{0.02}

(\sqrt{n})^2 = (\frac{2.327\sqrt{0.08*0.92}}{0.02})^2

n = 996.3

Rounding up:

A sample of 997 is needed.

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