1 A sample of 41 garages in a town has a mean of 192 cars wi

1. A sample of 41 garages in a town has a mean of 19.2 cars with a standard deviation of 3.5 trucks.

1a. Create a 99% confidence interval estimate for the standard deviation in number cars in this town. Use formula and chart.

1b. With .05 significance, test the claim that the standard deviation in number of cars is 5 trucks. Use the critical value (traditional) method.

1c. If the population standard deviation is known to be 3.8 trucks, test the claim that the mean number of cars in this town is greater than 18. Use the P-value method with formula and chart.

Solution

1a)sample size StDev Variance
41 1.87 3.50

99% Confidence Intervals

CI for CI for
Method StDev Variance
Chi-Square (1.45, 2.60) (2.10, 6.76)

1b)

Tests

Test
Method Statistic DF P-Value
Chi-Square 19.60 40 0.000

here the test statistics has chisq dustribution having df=40 for this we get lower alpha point of chi square is  26.5093, as test statistics value < critical value so we reject H0 : sigma=5

1c)

One-Sample Z

Test of mu = 18 vs > 18
The assumed standard deviation = 3.8

95% Lower
N Mean SE Mean Bound Z P
41 19.200 0.593 18.224 2.02 0.022

as p-value < 0.05 so we reject H0:mu = 18