Question

A soft drink manufacturer wishes to know how many soft drinks adults drink each week. They want to construct a 85% confidence interval with an error of no more than 0.06. A consultant has informed them that a previous study found the mean to be 7.6 soft drinks per week and found the variance to be 0.81. What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.

Answer 1

Answer:

The minimum sample size is 1867 observations.

Step-by-step explanation:

We need to construct an 85% confidence interval that has an error less than 0.06. It means that the difference between the upper limit (UL) and the lower limit (LL) has to be 0.06.

$UL-LL= e =0.06$

$UL-LL= e =0.06\\X+z*s/\sqrt{n} -(X-z*s/\sqrt{n}) = e\\2*z*s/\sqrt{n}=e$

The only variable we can adjust is the number of observations (n)

$2*z*s/\sqrt{n}=e\\\\\sqrt{n}=\frac{2*z*s}{e}\\\\ n=(\frac{2*z*s}{e})^{2}=\frac{4*z^{2} *s^{2} }{e^{2} }$

For a 85% confidence interval, the z-score is 1.440.

The estimated variance (s^2) is 0.81.

The error e is 0.06.

$n= 4*(1.440)^{2} *0.81/(0.06)^{2} \\\n=4*2.0736*0.81/0.0036= 6.7184 / 0.0036 = 1866.24$

The sample has to be at least of 1867 observations.