necessaryh

2021-01-05

The sample mean and population standard deviation are provided to you. Use this information to construct the $90\mathrm{\%}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}95\mathrm{\%}$ confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.

From a random sample of 66 dates, the mean record high daily temperature in a certain city has a mean of ${85.69}^{\circ}F$. Assume the population standard deviation is ${13.60}^{\circ}F.$

The $90\mathrm{\%}$ confidence interval is

The $95\mathrm{\%}$ confidence interval is

Which interval is wider?

Interpret the findings.

Derrick

Skilled2021-01-06Added 94 answers

Step 1

Given:

This mean recorded temperature is: $\stackrel{\u2015}{x}=85.69{F}^{\circ}$

The population's standard deviation is: $\sigma =13.60{F}^{\circ}$

The number of random samples is: $n=66$ dates.

Confidence level $-90\mathrm{\%}$:

The confidence level is expressed as,

$1-2\alpha =0.9$

$\alpha =\frac{1-0.9}{2}$

$=0.05$

Step 2

The confidence level can be seen to be symmetric about the P axis in the normal distribution graph. The formula to determine the appropriate value of P is,

$P=1-\alpha$

$=1-0.05$

$=0.95$

From the z-table corresponding to $P=0.95$, the value of z is 1.65

The expression to calculate the interval is,

$c.I=\stackrel{\u2015}{x}\pm z\frac{\sigma}{\sqrt{n}}$

Substitute the values in the above expression.

$c.l=85.69\pm 1.65\times \frac{13.60}{\sqrt{66}}$

$=85.69\pm 2.762$

$=(85.69-2.762,85.69-2.762)$

$=(82.928,88.452)$

Therefore, the interval for confidence leven $90\mathrm{\%}$ is from 82.928 to 88.452

Step 3

Confidence level $-95\mathrm{\%}:$

The expression for the confidence level is,

$1-2\alpha =0.95$

$\alpha =\frac{1-0.95}{2}$

$=0.025$

From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,

$P=1-\alpha$

$=1-0.025$

$=0.975$

From the z-table corresponding to $P=0.975$, the value of z is 1.96.

The expression to calculate the interval is,

$c.l=\stackrel{\u2015}{x}\pm z\frac{\sigma}{\sqrt{n}}$

Substitute the values in the above expression.

$c.l-85.69\pm 1.96\times \frac{13.60}{\sqrt{66}}$

$=85.69\pm 3.281$

$=(85.69-3.281,85.69-3.281)$

$=(82.409,88.971)$

Hence, the interval for confidence level $95\mathrm{\%}$ is from 82.409 to 88.971, and the interval for confidence level $95\mathrm{\%}$ is the widest one.

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