Question

In a sample of 49 individuals, it is found that the average amount spent on groceries per month is $255. From previous studies, it is assumed that the population standard deviation is $46. Find a 95% confidence interval for the average amount of money spent on groceries per month, and interpret your result.

Use ??=1.96 for 95% confidence, and round your answers to 2 decimal places.

The confidence interval is:

$

to

$

This means:

Select your answer from one of the following options.

• a.

  We can be 95% confident the population average for amount spent on groceries is between _____ and _____.

• b.

  We know for sure that the population mean for amount spent on groceries is between _____ and _____.

• c.

  95% of individuals spend between _____ and _____ on groceries.

• d.

  95% of the 49 individuals in the sample spent between _____ and _____ on groceries.

Answer 1

A) Given data

Sample size(n)=49

Mean Value (X')=255 $

Standard deviation (S)=46 $

We need to determine the 95% confidence interval

Z value=1.96

Formula used is

CI =X' ±Z×S/(√n)

=255 ± 1.96×46/(49)

=255 ±12.88

Lower limit=255-12.88=242.12

Upper limit=255+12.88=267.88

So 95% confidence interval is

(242.12 to 267.88)

Interpreting the confidence interval is

We can be 95% confident the population average for amount spent on groceries is between

242.12 and 267.88 (Option-A)