Question

A consumer advocacy group suspects that a local supermarket's 10 oz. packages of cheddar cheese actually weigh less than 10 oz. The standard deviation is 0.15 oz. The group took a random sample of 35 such packages and found that the mean weight for the sample was 9.95 oz. Calculate the confidence interval. Don't forget to label. Please go 4 decimal spaces on the error, but 2 on the boundaries.

Answer 1

Answer:  $(9.90,\ 10.00)$

Step-by-step explanation:

Confidence interval for population mean:

$\overline{x}\pm z^c\dfrac{\sigma}{\sqrt{n}}$

where , $\overline{x}$ = sample mean , $\sigma$ = population standard deviation, n= sample size, $z^c$ = Critical z-value.

As per given, $n= 35,\ \ \ \overline{x} =9.95,\ \ \ \ \sigma=0.15$

We will find 95% confidence interval for which $z^c=1.96$

Required confidence interval:

$9.95\pm(1.96)\dfrac{0.15}{\sqrt{35}} \\\\= 9.95\pm (0.049695)\\\\=( 9.95-0.049695,\ 9.95+0.049695)\\\\=(9.900305,\ 9.999695)\approx(9.90,\ 10.00)$

Required confidence interval : $(9.90,\ 10.00)$