Question

Let the following sample of 8 observations be drawn from a
normal population with unknown mean and standard deviation: 28, 23,
18, 15, 16, 5, 21, 13. **[You may find it useful to reference
the** t table**.]**

**a.** Calculate the sample mean and the sample
standard deviation. **(Round intermediate calculations to at
least 4 decimal places. Round "Sample mean" to 3 decimal places and
"Sample standard deviation" to 2 decimal places.)**

**b.** Construct the 90% confidence interval for
the population mean. **(Round " t" value to 3 decimal
places and final answers to 2 decimal places.)**

**c.** Construct the 95% confidence interval for
the population mean. **(Round " t" value to 3 decimal
places and final answers to 2 decimal places.)**

**d.** What happens to the margin of error as the
confidence level increases from 90% to 95%?

Answer #1

a) Since we know that

and n = 8

This implies that

b)

Confidence interval(in %) = 90

t = 1.8946

Since we know that

Required confidence interval = (17.375-4.6554, 17.375+4.6554)

Required confidence interval = (12.7196, 22.0304)

c)

Confidence interval(in %) = 95

t = 2.3646

Since we know that

Required confidence interval = (17.375-5.8103, 17.375+5.8103)

Required confidence interval = (11.5647, 23.1853)

d) when we increase the confidence interval, t value increases and so the value of margin of error is increased

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