Respuesta :
Answer:
[tex] 7.72 -2.539 \frac{2.5}{\sqrt{20}} =6.30[/tex]
[tex] 7.72 +2.539 \frac{2.5}{\sqrt{20}} =9.14[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex]\bar X= 7.72[/tex] represent the sample mean
[tex]s= 2.5[/tex] represent the sample deviation
[tex] n=20[/tex] represent the sample size
The confidence interval is given by:
[tex] \bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
The confidence interval is 98% and the significance level is [tex]\alpha=0.02[/tex] the degrees of freedom are given by:
[tex] df= n-1 = 20-1=19[/tex]
And the critical value would be:
[tex] t_{\alpha/2}= 2.539[/tex]
And replacing we got:
[tex] 7.72 -2.539 \frac{2.5}{\sqrt{20}} =6.30[/tex]
[tex] 7.72 +2.539 \frac{2.5}{\sqrt{20}} =9.14[/tex]
The 98% confidence interval is between 6.42 hamsters per liter to 9.02 hamsters per liter
Mean (μ) = 7.72, standard deviation (σ) = 2.5, sample size (n) = 20, Confidence (C) = 98% = 0.98
α = 1 - C = 0.02
α/2 = 0.01
The z score of α/2 is the same as the z score of 0.49 (0.5 - 0.01) which is equal to 2.326.
The margin of error E is:
[tex]E = Z_\frac{\alpha }{2} *\frac{\sigma}{\sqrt{n} } \\\\E=2.326*\frac{2.5}{\sqrt{20} } =1.3[/tex]
The confidence interval = (μ ± E) = (7.72 ± 1.3) = (6.42, 9.02)
Hence the 98% confidence interval is between 6.42 hamsters per liter to 9.02 hamsters per liter
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