Question

X 15.1.45 Find the 95% z-interval or t-interval for the indicated parameter. (a) u (b) p x = 153, s = 34, n=29 p=0.3, n= 60 (a) The 95% confidence interval for u is ( (Round to two decimal places as needed.) to

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = $\bar x$ = 153

sample standard deviation = s = 34

sample size = n = 29

Degrees of freedom = df = n - 1 = 29 -1 =28

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

t$\alpha$/2,df = t_0.025,28 = 2.048

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.048 * ( 34/ $\sqrt$ 29)

Margin of error = E = 12.93

The 95% confidence interval estimate of the population mean is,

$\bar x$  ± E

= 153 ± 12.93

= ( 140.07, 165.93 )

b) Given that,

n = 60

Point estimate = sample proportion = $\hat p$ = 0.3

1 - $\hat p$ = 1 - 0.3 = 0.7

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.960}

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.96 ($\sqrt$((0.3 * 0.7) / 60)

= 0.116

A 95% confidence interval for population proportion p is ,

$\hat p$   ± E

= 0.3  ± 0.116

= ( 0.184, 0.416 )