Answer: [tex]=(29.1596\text{ degrees},\ 30.2404\text{ degrees})[/tex]
Step-by-step explanation:
Given: Sample size: n = 41
Sample mean [tex]\overline{x}= 29.7[/tex] degrees
Population standard deviation [tex]\sigma=2.7[/tex] degrees
Confidence level (c) = 80% =0.80
Significance level (a) = 1- c = 1-0.80 = 0.20
z-score for 80% confidence level : z = 1.2816 [from z-table]
Confidence level for population mean :-
[tex]\overline{x}\pm z\dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]=29.7\pm ( 1.2816)\dfrac{2.7}{\sqrt{41}}[/tex]
[tex]=29.7\pm ( 1.2816)\dfrac{2.7}{6.403124}[/tex]
[tex]=29.7\pm ( 1.2816)(0.42167)[/tex]
[tex]=29.7\pm 0.5404[/tex]
[tex]=(29.7-0.5404,\ 29.7+0.5404)[/tex]
[tex]=(29.1596,\ 30.2404)[/tex]
Hence, 80% confidence interval for the temperatures in the freezer [tex]=(29.1596\text{ degrees},\ 30.2404\text{ degrees})[/tex]
