Respuesta :
Answer: ($3.532, $4.166)
Step-by-step explanation:
Given : Significance level : [tex]\alpha: 1-0.90=0.10[/tex]
Critical value : [tex]z_{\alpha/2}=1.645[/tex]
Sample size : n= 157
Sample mean : [tex]\overline{x}=\$\ 3.849[/tex]
Standard deviation : [tex]\sigma= \$\ 2.421[/tex]
The confidence interval for population mean is given by :_
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]\text{i.e. }\$\ 3.849\pm (1.64)\dfrac{2.421}{\sqrt{157}}\\\\\approx\$\ 3.849\pm0.317\\\\=(\$\ 3.849-0.317,\$\ 3.849+0.317)=(\$\ 3.532,\$\ 4.166)[/tex]
Hence, the 0% confidence interval for the average tip given at this restaurant = ($3.532, $4.166)
The confidence interval shows the the interval of values that will be achieved with chance = confidence percentage.
The confidence interval for 90% confidence for the average tip given at the considered restaurant is
[tex]CI \approx 3.849 \pm 0.193 \approx [3.66, 4.04][/tex]
How to calculate confidence interval for single mean for t statistic?
- If the sample size is given to be n,
- sample mean given as [tex]\overline{x}[/tex]
sample standard deviation s - The t statistic for given confidence interval and degree of freedom n-1, t
then
[tex]CI = \overline{x} \pm t\times \dfrac{s}{\sqrt{n}}[/tex]
For the given case, it is given that:
- n = 157
- degree of freedom = n - 1 = 156
- two tailed test (as no restriction on one side) and that price can be up or low
- s = 2.421
- [tex]\overline{x} = \$3.849[/tex]
- Confidence level = 90%
The critical value t at 156 degree of freedom for 90% confidence is 1.65
Thus, the confidence interval is given as
[tex]CI = \overline{x} \pm t\times \dfrac{s}{\sqrt{n}}\\\\CI = 3.849 \pm \dfrac{2.421}{\sqrt{157}}\\\\CI \approx 3.849 \pm 0.193 \approx [3.66, 4.04]\\[/tex]
Thus,
The confidence interval for 90% confidence for the average tip given at the considered restaurant is
[tex]CI \approx 3.849 \pm 0.193 \approx [3.66, 4.04][/tex]
Learn more about confidence interval here:
https://brainly.com/question/2396419
