Question

# Your client from question 4 decides they want a 90\% confidence interval instead of a 95\% one. You sampled 36 students to find out how many hours they spent online last week. Your sample mean is 46. The population standard deviation, \sigma, is 7.4. What is the 90\% confidence interval for this sample mean?

Confidence intervals
Your client from question 4 decides they want a $$\displaystyle{90}\%$$ confidence interval instead of a $$\displaystyle{95}\%$$ one. To remind you - you sampled 36 students to find out how many hours they spent online last week. Your sample mean is 46. The population standard deviation, $$\displaystyle\sigma$$, is 7.4. What is the $$\displaystyle{90}\%$$ confidence interval for this sample mean? Keep at least three decimal places on any intermediate calculations, then give your answer rounded to 1 decimal place, with the lower end first then upper.
(_______ , ___________ )

2021-08-16

Step 1
Given,
sample size $$\displaystyle{\left({n}\right)}={36}$$
sample mean $$\displaystyle{\left(\overline{{{x}}}\right)}={46}$$
standard deviation $$\displaystyle{\left(\sigma\right)}={7.4}$$
$$\displaystyle\alpha={1}-{0.90}={0.1}$$
$$\displaystyle{\frac{{\alpha}}{{{2}}}}={0.05}$$
$$\displaystyle{Z}_{{{0.05}}}={1.645}$$
Step 2
$$\displaystyle{90}\%$$ confidence interval $$\displaystyle{\left({C}.{I}.\right)}:\overline{{{X}}}\pm{Z}_{{{\frac{{\alpha}}{{{2}}}}}}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle{C}.{I}={46}\pm{1.645}\times{\frac{{{7.4}}}{{\sqrt{{{36}}}}}}$$
$$\displaystyle{C}.{I}={46}\pm{2.029}$$
$$\displaystyle{C}.{I}={\left({46}-{2.029},{46}+{2.029}\right)}$$
$$\displaystyle{C}.{I}={\left({43.971},{48.029}\right)}$$
$$\displaystyle{C}.{I}\stackrel{\sim}{=}{\left({44.0},{48.0}\right)}$$