A random sample of size 36 was taken from a population distributed as Nμ,3.92.The value of the sample x was 15.6. i. Find a 90% confidence interval for μ. (5mks)
It is believed that value of μ is 17. Use your confidence interval to comment on this belief.(2mks)
a)
CI = x ± z*"\\frac{\\sigma}{\\sqrt{n}}"
Given that x=15.6, "\\sigma"=3.92, n=36
From the normal distribution table, the critical value for 90% confidence interval z= ±1.645
Then;
90% CI = 15.6 ±1.96*"\\frac{3.92}{\\sqrt{36}}"
The lower limit = 15.6 - 1.28 =14.32
The upper limit = 15.6 + 1.28 = 16.88
Therefore, the 90% confidence interval for population mean is (14.32, 16.88)
b) Since the interval does not contain the value of "\\mu" =17, we can conclude that the value of the population mean "\\mu" "\\neq" 17
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