Question

3 Construct a 90% confidence interval for the following random sample of Lucas Barrett's golf scores for a particular golf course he played so that he can figure out his true (population) aver 95 92 95 99 92 84 95 94 95 86 (hint: Use T-distribution table. Formula Interval estimate of a population mean when stan age score for the dared deviation is unknown) 4 The auto industry relies heavily on 0% financing to entice customers to purchase cars. In 2017 it is estimated that 22.4% of car deals involved 0% financing. A random sample of 500 financed car deals found that 98 of them used 0% financing. (hint:sample size for an interval estimate of a population mean) Construct a 98% confidence interval around this sample proportion. Does it support the estimate?

Answer 1

3)   $\bar x$ = 92.7

S = 4.5228

At 90% confidence interval the critical value is t^* = 1.833

The 90% Confidence interval for population mean is

$\bar x$ +/- t^* * s/ $\sqrt n$

= 92.7 +/- 1.833 * 4.5228/ $\sqrt 10$

= 92.7 +/- 2.6216

= 90.0784, 95.3216

4)   $\widehat p$ = 98/500 = 0.196

At 98% Confidence interval the critical value is z^* = 2.33

The 98% Confidence interval for population proportion is

$\widehat p$ +/- z^* * sqrt( $\widehat p$ (1 - $\widehat p$ )/n)

= 0.196 +/- 2.33 * sqrt(0.196 * (1 - 0.196)/500)

= 0.196 +/- 0.041

= 0.155, 0.237

5) Margin of error = 5

Or, z_{0. 05} * $\sigma/\sqrt n$ = 5

Or, 1.645 * 32/ $\sqrt n$ = 5

Or, n = (1.645 * 32/5) ^2

Or, n = 111

6)   $\widehat p$ = 42/60 = 0.7

Margin of error = 0.06

Or, z_0.025 * sqrt( $\widehat p$ (1 - $\widehat p$ )/n) = 0.06

Or, 1.96 * sqrt(0.7 * (1 - 0.7)/n) = 0.06

Or, n = (1.96 * sqrt(0.7 * 0.3)/0.06)^2

Or, n = 225