Question

Suppose the weights, in pounds, of the dogs in a city are normally distributed. If the population standard deviation is 3 pounds, what minimum sample size is needed to be 95% confident that the sample mean is within 1 pound of the true population mean?

Answer 1

Answer: 35

Step-by-step explanation:

Given : Standard deviation : $\sigma=3\text{ pounds}$

Margin of error : $E=\pm1\text{ pound}$

Significance level : $\alpha: 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

The formula we use to find the minimum sample size required :-

$n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2$

i.e. $n=(\dfrac{(1.96)(3)}{1})^2=34.5744\approx35$

Hence, the minimum sample size is needed to be 95% confident that the sample mean is within 1 pound of the true population mean =35

Answer 2

Answer:

35 dogs

Step-by-step explanation:

Correct Answer 35 dogs

The formula for sample size is

n=z2σ2EBM2

In this formula,

z=zα2=z0.025=1.96

because the confidence level is 95%. From the problem, we know that σ=3 and EBM=1. Therefore,

n=z2σ2EBM2=(1.96)2(3)212≈34.57

Use n=35 to ensure that the sample size is large enough.

Also, the sample size formula shown above is sometimes written using an alternate format of n=(zσE)2. In this formula, E is used to denote margin of error and the entire parentheses is raised to the exponent 2.

Therefore, the margin of error for the mean can be denoted by "EBM" or by "E". Either formula for the sample size can be used and these formulas are considered as equivalent.