Question

A random sample of a specific brand of snack bar is tested for calorie count, with the following results: Assume the population standard deviation is σ and that the population is approximately normal. Construct a 95% confidence interval for the calorie count of the snack bars.

Answer 1

Answer:

The 95% confidence interval for the population mean (calorie count of the snacks bars) is (130.32, 161.68).

Step-by-step explanation:

The question is incomplete:

"A random sample of a specific brand of snack bar is tested for calorie count, with the following results: 149, 145,140,160,149,153,131,134,153. Assume the population standard deviation is σ=24 and that the population is approximately normal. Construct a 95% confidence interval for the calorie count of the snack bars."

We start by calculating the mean of the sample:

$M=\dfrac{1}{9}\sum_{i=1}^{9}(149+145+140+160+149+153+131+134+153)\\\\\\ M=\dfrac{1314}{9}=146$

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is know and is σ=24.

The sample mean is M=146.

The sample size is N=9.

As σ is known, the standard error of the mean (σM) is calculated as: $\sigma_M=\dfrac{\sigma}{\sqrt{N}}=\dfrac{24}{\sqrt{9}}=\dfrac{24}{3}=8$

The z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_M=1.96 \cdot 8=15.68$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 146-15.68=130.32\\\\UL=M+t \cdot s_M = 146+15.68=161.68$

The 95% confidence interval for the population mean is (130.32, 161.68).