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

a)
`\bar X = (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

Explanation:

Part a

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

`\bar X = (\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 = (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)(s)/(√(n))` (1)

The margin of error is given by this formula:

`ME=t_(\alpha/2)(s)/(√(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