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Sagot :
Answer:
98% confidence interval for the population mean =(1095.260,1288.740)
Step-by-step explanation:
We are given that
n=45
[tex]\mu=1192[/tex]
Standard deviation,[tex]\sigma=279[/tex]
We have to construct a 98% confidence interval for the population mean.
Critical value of z at 98% confidence, Z =2.326
Confidence interval is given by
[tex](\mu\pm Z\frac{\sigma}{\sqrt{n}})[/tex]
Using the formula
98% confidence interval is given by
[tex]=(1192\pm 2.326\times \frac{279}{\sqrt{45}})[/tex]
[tex]=(1192\pm 96.740)[/tex]
=[tex](1192-96.740,1192+96.740)[/tex]
=[tex](1095.260,1288.740)[/tex]
Hence, 98% confidence interval for the population mean (1095.260,1288.740)
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