Question

In response to a survey question about the number of hours daily spent watching TV, the responses by the eight subjects who identified themselves as Hindu were 2 , 2 , 1 , 3 , 1 , 0 , 4 , 1
a. Find a point estimate of the population mean for Hindus.

--------------(Round to two decimal places as needed)

b. The margin of error at the 95% confidence level for this point estimate is 0.89. Explain what this represents.

The margin of error indicates we can be__%confident that the sample mean falls within __ of the _____(population mean/ standard error/ sample mean)

Answer 1

Answer:

a) $\bar X = \frac{2+2+1+3+1+0+4+1}{8}= 1.75$

b) The margin of error indicates we can be 95%confident that the sample mean falls within 0.89 of the population mean

Step-by-step explanation:

Part a

The best point of estimate for the population mean is the sample mean given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Since is an unbiased estimator $E(\bar X) = \mu$

Data given: 2 , 2 , 1 , 3 , 1 , 0 , 4 , 1

So for this case the sample mean would be:

$\bar X = \frac{2+2+1+3+1+0+4+1}{8}= 1.75$

Part b

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The margin of error is given by this formula:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$    (2)

And for this case we know that ME =0.89 with a confidence of 95%

So then the limits for our confidence level are:

$Lower= \bar X -ME= 1.75- 0.89=0.86$

$Upperr= \bar X +ME= 1.75+0.89=2.64$

So then the best answer for this case would be:

The margin of error indicates we can be 95%confident that the sample mean falls within 0.89 of the population mean