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3 Construct a 90% confidence interval for the following random sample of Lucas Barretts 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?
5 The park rangers at a hiking trail would like to estimate the mean hiking time for visitors along a specific trail, a five-mile roundtrip trek from the parking area. Assume the population standard deviation for hiking the trail is 32 minutes Determine the sample size needed to construct a 90% confidence interval to estimate the mean hiking time with a margin of error within+ or -5 minutes 6 The federal government would like to estimate the proportion of Americans who own their own homes A sample of 60 Americans found that 42 were homeowners. Determine the sample size required to construct a 95% confidence interval to estimate the proportion of homeowners when the margin of error equals 6%. (hint: sample size for4 an interval estimate of a population proportion)
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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, z0. 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, z0.025 * 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

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