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A sample of size 168, taken from a normally distributed population whose standard deviation is known to be 8.60, has a sample mean of 80.60. Suppose that we have adopted the null hypothesis that the actual population mean is equal to 80, that is, H0 is that μ = 80 and we want to test the alternative hypothesis, H1, that μ ≠ 80, with level of significance α = 0.1. The upper limit of a 95% confidence interval for the population mean would equal: _______

Answer :

Answer:

The 95% confidence interval for the population mean is

(79.2996, 81.9004)

Step-by-step explanation:

Step(i):-

Given that the sample size 'n' = 168

Let 'X' be a Random variable in a normal distribution

Given that mean of the sample x⁻ = 80.60

Given that the standard deviation of the Population = 8.60

Step(ii):-

The 95% confidence interval for the population mean is determined by

[tex](x^{-} - Z_{0.05} \frac{S.D}{\sqrt{n} } ,x^{-} + Z_{0.05} \frac{S.D}{\sqrt{n} } )[/tex]

[tex](80.60 -1.96 \frac{8.60}{\sqrt{168} } ,80.60 + 1.96 \frac{8.60}{\sqrt{168} } )[/tex]

(80.60 -1.3004 , 80.60+1.3004)

(79.2996 , 81.9004)

Final answer:-

The 95% confidence interval for the population mean is

(79.2996, 81.9004)

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