Question

If we are very certain that the true standard deviation of the number of defects per screen is below 2, what sample size would be required so that the width of the 95% confidence interval for the mean number of defects per screen is at most 0.27? Make sure you enter a whole number below.

Answer 1

Answer: 211

Step-by-step explanation:

As per given , we have

The true standard deviation: $\sigma<2$

Critical value for 95% confidence = $z_{\alpha/2}=1.96$

margin of error : E= 0.27

Formula to find the required sample size : -

$n=(\dfrac{z_{\alpha/2}\sigma}{E})^2<(\dfrac{1.96\times2}{0.27})^2\\\\=210.787379973\approx211$

hence, the required minimum sample =211

Answer 2

A sample size of 211 would be required so that the width of the 95% confidence interval for the mean number of defects per screen is at most 0.27

What is margin of error?

Margin of error is used to determine by what value there is deviation from the real value. Margin of error (E) is given by:

$E=Z_\frac{\alpha}{2} *\frac{standard\ deviation}{\sqrt{sample\ size} }$

The 95% confidence level have a z score of 1.96. Hence for E = 0.27, standard deviation = 2 , hence:

$0.27=1.96*\frac{2}{\sqrt{sample\ size}}\\ \\sample\ size=211$

A sample size of 211 would be required so that the width of the 95% confidence interval for the mean number of defects per screen is at most 0.27

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