Answer to Question #140005 in Statistics and Probability for Sohyun Ravena

Question

~Include a normal curve/t distribution curve.
~Your final answer must be a concluding statement.

To determine the views of students on a particular campus about palabok, an 11-point attitude scale was administered to a random sample of 40 students. This survey yielded a sample of 6 and a standard deviation of 1.5
1. Compute for the standard error of the mean.
2. Find the 95% confidence interval for the population mean.
3. Find the 99% confidence interval for the population mean.

Answer 1

The provided sample mean is $\bar X = 6$

 and the sample standard deviation is s = 1.5. The size of the sample is n = 40 and the required confidence level is 95%.

1) standard error= $\frac{tc*s}{\sqrt{n}}$

Based on the provided information, the critical t-value for α=0.05 and df = n-1= 40-1=39

is tc​=2.023

so standard error= $\frac{2.023*1.5}{\sqrt{40}}$ =0.48

ii)The 95% confidence for the population mean $\mu$ is computed using the following expression

$[\overline{X} - \frac{tc*s}{\sqrt{n}},\overline{X} +\frac{tc*s}{\sqrt{n}}]$

Substitute the values ,

$[6 - \frac{2.023*1.5}{\sqrt{40}},6 + \frac{2.023*1.5}{\sqrt{40}}]$

=(6−0.48,6+0.48)

=(5.52,6.48)

iii)Based on the provided information, the critical t-value for α=0.01 and df = n-1= 40-1=39 is tc​=2.708

so confidence interval is given by

$[\overline{X} - \frac{tc*s}{\sqrt{n}},\overline{X} +\frac{tc*s}{\sqrt{n}}]$

$[6 - \frac{2.708*1.5}{\sqrt{40}},6 + \frac{2.708*1.5}{\sqrt{40}}]$

=(6−0.642,6+0.642)

=(5.358,6.642).