Question

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11.* A random sample of size n 64 is drawn from a population with mean μ and standard deviation σ. The mean and standard deviation of the sample are X = 308.9 and s 31.9 a. Find a 90%confidence interval for the mean μ. Interpret this interval. b. Find a 95%confidence interval for the mean μ. Interpret this interval. c. Find a 99%confidence interval for the mean μ. Interpret this interval. d. Compare the widths of the three confidence intervals. Which interval is the widest? The narrowest?

Answer 1

This is the case of finding the confidence interval for the mean $\mu$ when $\sigma^2$ is unknown,

and this confidence interval is given by,

$P[\bar{x}-\frac{s}{\sqrt{n}}t_{\alpha/2}\leq \mu \leq \bar{x}+\frac{s}{\sqrt{n}}t_{\alpha/2}]=1-\alpha$

where,

$\bar{x}=sample\quad mean$

$s=sample\quad standard \quad deviation$

$n=total\quad observations$

we are given that,

$\bar{x}=308.9$

$s=31.9$ and

$n=64$

therefore $100(1-\alpha)%$ % is given by,

$P[308.9-\frac{31.9}{\sqrt{64}}t_{\alpha/2}\leq \mu \leq 308.9+\frac{31.9}{\sqrt{64}}t_{\alpha/2}]=1-\alpha$

(a).

for a 90% confidence interval $\alpha=0.1$

$\therefore t_{\alpha/2}=t_{0.05}=1.671$ [value generated from the t-table]

hence the required confidence interval is given by,

$[308.9-\frac{31.9}{8}*1.671,308.9+\frac{31.9}{8}*1.671]$

$=[308.9-6.663,308.9+6.663]$

$=[302.237,315.563]$

That is we can say with 90% confidence that the true value of the mean lies between

(b).

for a 95% confidence interval $\alpha=0.05$

$\therefore t_{\alpha/2}=t_{0.025}=2$ [value generated from the t-table]

hence the required confidence interval is given by,

$[308.9-\frac{31.9}{8}*2,308.9+\frac{31.9}{8}*2]$

$=[308.9-7.975,308.9+7.975]$

$=[300.925,316.875]$

That is we can say with 95% confidence that the true value of the mean lies between

(c).

for a 99% confidence interval $\alpha=0.01$

$\therefore t_{\alpha/2}=t_{0.005}=2.660$ [value generated from the t-table]

hence the required confidence interval is given by,

$[308.9-\frac{31.9}{8}*2.66,308.9+\frac{31.9}{8}*2.66]$

$=[308.9-10.61,308.9+10.61]$

$=[298.29,319.51]$

That is we can say with 99% confidence that the true value of the mean lies between

(d).

Now comparing the three confidence interval

a -

b -

c -

the widest interval is the 99% confidence interval that is (c) -

and the narrowest interval is the 90% confidence interval that is (a) -